Properties of Triangles

In $\triangle ABC$, prove that $4 \left[b c \cos^2 \left(\dfrac{A}{2}\right) + c a \cos^2 \left(\dfrac{B}{2}\right) + a b \cos^2 \left(\dfrac{C}{2}\right)\right] = \left(a + b + c\right)^2$

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\begin{aligned} LHS & = 4 \left[b c \cos^2 \left(\dfrac{A}{2}\right) + c a \cos^2 \left(\dfrac{B}{2}\right) + a b \cos^2 \left(\dfrac{C}{2}\right)\right] \\\\ & \left[\text{Note: } \cos^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 + \cos \theta}{2}\right] \\\\ & = 4 \left[b c \left(\dfrac{1 + \cos A}{2}\right) + c a \left(\dfrac{1 + \cos B}{2}\right) + a b \left(\dfrac{1 + \cos C}{2}\right)\right]\\\\ & = 2 \left[b c \left(1 + \dfrac{b^2 + c^2 - a^2}{2 b c}\right) + c a \left(1 + \dfrac{c^2 + a^2 - b^2}{2 c a}\right) \right. \\\\ & \left. \hspace{3cm} + a b \left(1 + \dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] \;\; \left[\text{by cosine rule}\right] \\\\ & = 2 b c + b^2 + c^2 - a^2 + 2 c a + c^2 + a^2 - b^2 + 2 a b + a^2 + b^2 - c^2 \\\\ & = a^2 + b^2 + c^2 + 2 a b + 2 b c + 2 c a \\\\ & = \left(a + b + c\right)^2 = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $2 \left[a \sin^2 \left(\dfrac{C}{2}\right) + c \sin^2 \left(\dfrac{A}{2}\right) \right] = c + a - b$

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\begin{aligned} LHS & = 2 \left[a \sin^2 \left(\dfrac{C}{2}\right) + c \sin^2 \left(\dfrac{A}{2}\right) \right] \\\\ & \left[\text{Note: } \sin^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{2}\right] \\\\ & = 2 \left[a \left(\dfrac{1 - \cos C}{2}\right) + c \left(\dfrac{1 - \cos A}{2}\right)\right] \\\\ & = a \left[1 - \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] + c \left[1 - \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right)\right] \;\;\; \left[\text{by cosine rule}\right] \\\\ & = a \left(\dfrac{2 a b - a^2 - b^2 + c^2}{2 a b}\right) + c \left(\dfrac{2 b c - b^2 - c^2 + a^2}{2 b c}\right) \\\\ & = \dfrac{1}{2 b} \left(2 a b - a^2 - b^2 + c^2 + 2 b c - b^2 - c^2 + a^2\right) \\\\ & = \dfrac{1}{2 b} \left(2 a b + 2 b c - 2 b^2\right) \\\\ & = a + c - b = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $\left(\dfrac{b^2 - c^2}{a^2}\right) \sin 2A + \left(\dfrac{c^2 - a^2}{b^2}\right) \sin 2B + \left(\dfrac{a^2 - b^2}{c^2}\right) \sin 2 C = 0$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

Now,

$\begin{aligned} \left(\dfrac{b^2 - c^2}{a^2}\right) \sin 2 A & = \left(\dfrac{b^2 - c^2}{a^2}\right) \times 2 \sin A \cos A \\\\ & = \left(\dfrac{b^2 - c^2}{a^2}\right) \times 2 \times \dfrac{a}{2 R} \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) \\ & \hspace{3cm} \left[\text{by sine and cosine rules}\right] \\\\ & = \dfrac{\left(b^2 - c^2\right) \left(b^2 + c^2 - a^2\right)}{2 R a b c} \\\\ & = \dfrac{b^4 + b^2 c^2 - b^2 a^2 - b^2 c^2 - c^4 + a^2 c^2}{2 R a b c} \\\\ & = \dfrac{b^4 - b^2 a^2 - c^4 + a^2 c^2}{2 R a b c} \;\;\; \cdots \; (1a) \end{aligned}$

Similarly,

$\begin{aligned} \left(\dfrac{c^2 - a^2}{b^2}\right) \sin 2 B & = \left(\dfrac{c^2 - a^2}{b^2}\right) \times 2 \sin B \cos B \\\\ & = \dfrac{c^4 - b^2 c^2 -a^4 + a^2 b^2}{2 R a b c} \;\;\; \cdots \; (1b) \end{aligned}$

$\begin{aligned} \left(\dfrac{a^2 - b^2}{c^2}\right) \sin 2 C & = \left(\dfrac{a^2 - b^2}{c^2}\right) \times 2 \sin C \cos C \\\\ & = \dfrac{a^4 - a^2 c^2 - b^4 + b^2 c^2}{2 R a b c} \;\;\; \cdots \; (1c) \end{aligned}$

Adding equations $(1a)$, $(1b)$ and $(1c)$ we have,

$\begin{aligned} LHS & = \left(\dfrac{b^2 - c^2}{a^2}\right) \sin 2A + \left(\dfrac{c^2 - a^2}{b^2}\right) \sin 2B + \left(\dfrac{a^2 - b^2}{c^2}\right) \sin 2 C \\\\ & = \dfrac{1}{2 R a b c} \left(b^4 - a^2 b^2 - c^4 + a^2 c^2 \right. \\\\ & \left. \hspace{2.5cm} + c^4 - b^2 c^2 - a^4 + a^2 b^2 \right. \\\\ & \left. \hspace{3.5cm} + a^4 - a^2 c^2 - b^4 + b^2 c^2 \right) \\\\ & = 0 = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $\left(b^2 - c^2\right) \cot A + \left(c^2 - a^2\right) \cot B + \left(a^2 - b^2\right) \cot C = 0$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

Now,

$\begin{aligned} \left(b^2 - c^2\right) \cot A & = \left(4 R^2 \sin^2 B - 4 R^2 \sin^2 C\right) \times \dfrac{\cos A}{\sin A} \;\;\; \left[\text{by sine rule}\right] \\\\ & = 4 R^2 \left(\sin^2 B - \sin^2 C\right) \times \dfrac{\cos A}{\sin A} \\\\ & \left[\text{Note: } \sin^2 \alpha - \sin^2 \beta = \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta \right)\right] \\\\ & = 4 R^2 \sin \left(B + C\right) \sin \left(B - C\right) \times \dfrac{\cos A}{\sin A} \\\\ & \left[\text{In } \triangle ABC, \; A + B + C = \pi \right. \\\\ & \left. \therefore \; B + C = \pi - A \right. \\\\ & \left. \therefore \; \sin \left(B + C\right) = \sin \left(\pi - A\right) = \sin A \right] \\\\ & = 4 R^2 \sin A \sin \left(B - C\right) \times \dfrac{\cos A}{\sin A} \\\\ & = 4 R^2 \sin \left(B - C\right) \cos A \\\\ & = 4 R^2 \left(\sin B \cos C - \cos B \sin C\right) \cos A \\\\ & = 4 R^2 \left(\cos A \sin B \cos C - \cos A \cos B \sin C\right) \;\;\; \cdots \; (1a) \end{aligned}$

Similarly,

$\begin{aligned} \left(c^2 - a^2\right) \cot B & = \left(4 R^2 \sin^2 C - 4 R^2 \sin^2 A\right) \times \dfrac{\cos B}{\sin B} \\\\ & = 4 R^2 \left(\cos A \cos B \sin C - \sin A \cos B \cos C\right) \;\;\; \cdots \; (1b) \end{aligned}$

$\begin{aligned} \left(a^2 - b^2\right) \cot C & = \left(4 R^2 \sin^2 A - 4 R^2 \sin^2 B\right) \times \dfrac{\cos C}{\sin C} \\\\ & = 4 R^2 \left(\sin A \cos B \cos C - \cos A \sin B \cos C\right) \;\;\; \cdots \; (1c) \end{aligned}$

Adding equations $(1a)$, $(1b)$ and $(1c)$ we have,

$\begin{aligned} LHS & = \left(b^2 - c^2\right) \cot A + \left(c^2 - a^2\right) \cot B + \left(a^2 - b^2\right) \cot C \\\\ & = 4 R^2 \left[ \cos A \sin B \cos C - \cos A \cos B \sin C \right. \\\\ & \left. \hspace{1.5cm} + \cos A \cos B \sin C - \sin A \cos B \cos C \right. \\\\ & \left. \hspace{2cm} + \sin A \cos B \cos C - \cos A \sin B \cos C \right] \\\\ & = 0 = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $\dfrac{1 + \cos A \cos \left(B - C\right)}{1 + \cos C \cos \left(A - B\right)} = \dfrac{b^2 + c^2}{b^2 + a^2}$

$\left\{\text{Note: } \cos \alpha \cos \beta = \dfrac{1}{2} \left[\cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right)\right] \right\}$

$\begin{aligned} LHS & = \dfrac{1 + \cos A \cos \left(B - C\right)}{1 + \cos C \cos \left(A - B\right)} \\\\ & = \dfrac{1 + \dfrac{1}{2} \left[\cos \left(A + B - C\right) + \cos \left(A - B + C\right)\right]}{1 + \dfrac{1}{2} \left[\cos \left(C + A - B\right) + \cos \left(C - A + B\right)\right]} \\\\ & = \dfrac{2 + \cos \left(A + B - C\right) + \cos \left(A - B + C\right)}{2 + \cos \left(C + A - B\right) + \cos \left(C - A + B\right)} \;\;\; \cdots \; (1) \end{aligned}$

In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$

$\therefore \;$ $A + B = \pi - C$ $\;\; \cdots \; (2a)$,

$A + C = \pi - B$ $\;\; \cdots \; (2b)$,

$B + C = \pi - A$ $\;\; \cdots \; (2c)$

$\therefore \;$ In view of equations $(2a)$, $(2b)$ and $(2c)$ equation $(1)$ becomes,

$\begin{aligned} LHS & = \dfrac{2 + \cos \left(\pi - C - C\right) + \cos \left(\pi - B - B\right)}{2 + \cos \left(\pi - B - B\right) + \cos \left(\pi - A - A\right)} \\\\ & = \dfrac{2 + \cos \left(\pi - 2 C\right) + \cos \left(\pi - 2 B\right)}{2 + \cos \left(\pi - 2 B\right) + \cos \left(\pi - 2 A\right)} \\\\ & \left[\text{Note: } \cos \left(\pi - \theta\right) = - \cos \theta\right] \\\\ & = \dfrac{2 - \cos \left(2 C\right) - \cos \left(2 B\right)}{2 - \cos \left(2 B\right) - \cos \left(2 A\right)} \\\\ & \left[\text{Note: } \cos \left(2 \theta\right) = 1 - 2 \sin^2 \theta\right] \\\\ & = \dfrac{2 - \left(1 - 2 \sin^2 C\right) - \left(1 - 2 \sin^2 B\right)}{2 - \left(1 - 2 \sin^2 B\right) - \left(1 - 2 \sin^2 A\right)} \\\\ & = \dfrac{2 - 1 + 2 \sin^2 C - 1 + 2 \sin^2 B}{2 - 1 + 2 \sin^2 B - 1 + 2 \sin^2 A} \\\\ & = \dfrac{\sin^2 C + \sin^2 B}{\sin^2 B + \sin^2 A} \;\;\; \cdots \; (3) \end{aligned}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $\sin A = \dfrac{a}{2 R}$ $\;\; \cdots \; (4a)$, $\;$ $\sin B = \dfrac{b}{2 R}$ $\;\; \cdots \; (4b)$, $\;$ $\sin C = \dfrac{c}{2 R}$ $\;\; \cdots \; (4c)$

$\therefore \;$ In view of equations $(4a)$, $(4b)$ and $(4c)$ equation $(3)$ becomes,

$\begin{aligned} LHS & = \dfrac{\dfrac{c^2}{4 R^2} + \dfrac{b^2}{4 R^2}}{\dfrac{b^2}{4 R^2} + \dfrac{a^2}{4 R^2}} \\\\ & = \dfrac{c^2 + b^2}{b^2 + a^2} \\\\ & = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $\tan \left(\dfrac{A}{2} + B\right) = \left(\dfrac{c + b}{c - b}\right) \tan \left(\dfrac{A}{2}\right)$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

$\begin{aligned} RHS & = \left(\dfrac{c + b}{c - b}\right) \tan \left(\dfrac{A}{2}\right) \\\\ & = \left(\dfrac{2 R \sin C + 2 R \sin B}{2 R \sin C - 2 R \sin B}\right) \tan \left(\dfrac{A}{2}\right) \;\; \left[\text{by sine rule}\right] \\\\ & = \left(\dfrac{\sin C + \sin B}{\sin C - \sin B}\right) \tan \left(\dfrac{A}{2}\right) \\\\ & \left\{\text{Note: } \sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right) \right\} \\\\ & \left\{\text{Note: } \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right) \right\} \\\\ & = \dfrac{2 \sin \left(\dfrac{C + B}{2}\right) \cos \left(\dfrac{C - B}{2}\right)}{2 \sin \left(\dfrac{C - B}{2}\right) \cos \left(\dfrac{C + B}{2}\right)} \times \dfrac{\sin \left(\dfrac{A}{2}\right)}{\cos \left(\dfrac{A}{2}\right)} \;\;\; \cdots \; (1) \end{aligned}$

In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $B + C = \pi - A$

i.e. $\;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$

$\therefore \;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (2a)$

and $\;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (2b)$

$\therefore \;$ In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,

$\begin{aligned} RHS & = \dfrac{\cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{C - B}{2}\right) \sin \left(\dfrac{A}{2}\right)}{\sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{C - B}{2}\right) \cos \left(\dfrac{A}{2}\right)} \\\\ & = \dfrac{\cos \left(\dfrac{C - B}{2}\right)}{\sin \left(\dfrac{C - B}{2}\right)} \\\\ & = \cot \left(\dfrac{C - B}{2}\right) \;\;\; \cdots \; (3) \end{aligned}$

Now, in $\;$ $\triangle ABC$, $\;$ $C = \pi - A - B$

i.e. $\;$ $C - B = \pi - A - 2 B$

i.e. $\;$ $\dfrac{C - B}{2} = \dfrac{\pi}{2} - \left(\dfrac{A}{2} + B\right)$

$\therefore \;$ $\cot \left(\dfrac{C - B}{2}\right) = \cot \left[\dfrac{\pi}{2} - \left(\dfrac{A}{2} + B\right)\right] = \tan \left(\dfrac{A}{2} + B\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ In view of equation $(4)$ equation $(3)$ becomes,

$RHS = \tan \left(\dfrac{A}{2} + B\right) = LHS$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $\sin \left(\dfrac{A - B}{2}\right) = \left(\dfrac{a - b}{c}\right) \cos \left(\dfrac{C}{2}\right)$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $C = \pi - \left(A + B\right)$

$\therefore \;$ $\dfrac{C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right)$

$\therefore \;$ $\cos \left(\dfrac{C}{2}\right) = \cos \left[\dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right)\right] = \sin \left(\dfrac{A + B}{2}\right)$ $\;\;\; \cdots \; (1)$

$\begin{aligned} RHS & = \left(\dfrac{a - b}{c}\right) \cos \left(\dfrac{C}{2}\right) \\\\ & = \left(\dfrac{2 R \sin A - 2 R \sin B}{2 R \sin C}\right) \sin \left(\dfrac{A + B}{2}\right) \;\; \left[\text{by sine rule and equation } (1)\right] \\\\ & = \left(\dfrac{\sin A - \sin B}{\sin C}\right) \sin \left(\dfrac{A + B}{2}\right) \\\\ & = \dfrac{1}{\sin C} \times 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) \times \sin \left(\dfrac{A + B}{2}\right) \\\\ & \left[\text{Note: } 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) = \sin \left(A + B\right)\right] \\\\ & = \dfrac{1}{\sin C} \times \sin \left(\dfrac{A - B}{2}\right) \times \sin \left(A + B\right) \;\;\; \cdots \; (2) \end{aligned}$

In $\;$ $\triangle ABC$, $\;$ $A + B = \pi - C$

$\therefore \;$ $\sin \left(A + B\right) = \sin \left(\pi - C\right) = \sin C$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of equation $(3)$, equation $(2)$ becomes

$\begin{aligned} RHS & = \dfrac{1}{\sin C} \times \sin \left(\dfrac{A - B}{2}\right) \times \sin C \\\\ & = \sin \left(\dfrac{A - B}{2}\right) = LHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $a \sin \left(\dfrac{A}{2} + C\right) = \left(b + c\right) \sin \left(\dfrac{A}{2}\right)$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

$\begin{aligned} RHS = \left(b + c\right) \sin \left(\dfrac{A}{2}\right) & = b \sin \left(\dfrac{A}{2}\right) + c \sin \left(\dfrac{A}{2}\right) \\\\ & = 2R \sin B \sin \left(\dfrac{A}{2}\right) + 2 R \sin C \sin \left(\dfrac{A}{2}\right) \; \left[\text{by sine rule}\right] \\\\ & = 2 R \sin \left(\dfrac{A}{2}\right) \left[\sin B + \sin C\right] \\\\ & = 2 R \sin \left(\dfrac{A}{2}\right) \times 2 \sin \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \;\;\; \cdots \; (1) \end{aligned}$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

$\therefore \;$ $B + C = \pi - A$

$\therefore \;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$

$\therefore \;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes

$\begin{aligned} RHS & = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \\\\ & = 2 R \times \left[2 \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{A}{2}\right)\right] \times \cos \left(\dfrac{B - C}{2}\right) \\\\ & = 2 R \sin A \cos \left(\dfrac{B - C}{2}\right) \\\\ & = a \cos \left(\dfrac{B - C}{2}\right) \;\;\; \left[\text{By sine rule}\right] \;\;\; \cdots \; (3) \end{aligned}$

$\because \;$ In $\triangle ABC$, $\;$ $A + B + C = \pi$

$\therefore \;$ $B = \pi - A - C$

$\therefore \;$ $B = \pi - A - 2 C$

$\therefore \;$ $\dfrac{B - C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A}{2} + C\right)$

$\therefore \;$ $\cos \left(\dfrac{B - C}{2}\right) = \cos \left[\dfrac{\pi}{2} - \left(\dfrac{A}{2} + C\right)\right] = \sin \left(\dfrac{A}{2} + C\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ In view of equation $(4)$ equation $(3)$ becomes

$RHS = a \sin \left(\dfrac{A}{2} + C\right) = LHS$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $a^2 \left(\cos^2 B - \cos^2 C\right) + b^2 \left(\cos^2 C - \cos^2 A\right) + c^2 \left(\cos^2 A - \cos^2 B\right) = 0$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

Now, $\;$ $\cos^2 B - \cos^2 C $

$ = \left(\cos B + \cos C\right) \left(\cos B - \cos C\right) $

$ = 2 \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \times \left(-2\right) \sin \left(\dfrac{B + C}{2}\right) \sin \left(\dfrac{B - C}{2}\right) $

In $\triangle ABC$, $\;$ $B + C = \pi - A$

i.e. $\;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$

$\therefore \;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin \left(\dfrac{A}{2}\right)$

and, $\;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$

$\therefore \;$ $\cos^2 B - \cos^2 C$

$= - 2 \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \times 2 \cos \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B - C}{2}\right)$

$= - \sin A \sin \left(B - C\right)$ $\;\;\; \cdots \; (1a)$

Similarly, $\;$ $\cos^2 C - \cos^2 A = - \sin B \sin \left(C - A\right)$ $\;\;\; \cdots \; (1b)$

and $\;$ $\cos^2 A - \cos^2 B = - \sin C \sin \left(A - B\right)$ $\;\;\; \cdots \; (1c)$

$\therefore \;$ In view of equations $(1a)$, $(1b)$ and $(1c)$ we have,

$LHS = a^2 \left(\cos^2 B - \cos^2 C\right) + b^2 \left(\cos^2 C - \cos^2 A\right) + c^2 \left(\cos^2 A - \cos^2 B\right)$

$= - a^2 \sin A \sin \left(B - C\right) - b^2 \sin B \sin \left(C - A\right) - c^2 \sin C \sin \left(A - B\right)$ $\;\;\; \cdots \; (2)$

$\therefore \;$ Using sine rule, equation $(2)$ can be written as

$LHS = - \dfrac{1}{2 R} \left\{ a^3 \sin \left(B - C\right) + b^3 \sin \left(C - A\right) + c^3 \sin \left(A - B\right) \right\}$

$= - \dfrac{1}{2 R} \left\{a^3 \left(\sin B \cos C - \cos B \sin C\right) \right.$

$\left. \hspace{2cm} + b^3 \left(\sin C \cos A - \cos C \sin A\right) \right.$

$\left. \hspace{3cm} + c^3 \left(\sin A \cos B - \cos A \sin B\right) \right\}$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Using sine rule and cosine rule, equation $(3)$ can be written as

$LHS = \dfrac{-1}{2R} \left\{a^3 \left[\left(\dfrac{b}{2R}\right) \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) - \left(\dfrac{c}{2R}\right) \times \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right)\right] \right.$

$\left. \hspace{2cm} + b^3 \left[\left(\dfrac{c}{2R}\right) \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) - \left(\dfrac{a}{2R}\right) \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] \right.$

$\left. \hspace{3cm} + c^3 \left[\left(\dfrac{a}{2R}\right) \times \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) - \left(\dfrac{b}{2R}\right) \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right)\right] \right\}$

$= \dfrac{-1}{8R^2} \left\{a^2 \left[\left(a^2 + b^2 - c^2\right) - \left(c^2 + a^2 - b^2\right)\right] \right.$

$\left. \hspace{2.5cm} + b^2 \left[\left(b^2 + c^2 - a^2\right) - \left(a^2 + b^2 - c^2\right)\right] \right.$

$\left. \hspace{3.5cm} + c^2 \left[\left(c^2 + a^2 - b^2\right) - \left(b^2 + c^2 - a^2\right)\right] \right\}$

$= \dfrac{-1}{8R^2} \left\{a^2 \left(2 b^2 - 2 c^2\right) + b^2 \left(2 c^2 - 2 a^2\right) + c^2 \left(2 a^2 - b^2\right) \right\}$

$= \dfrac{-1}{4 R^2} \left\{a^2 b^2 - a^2 c^2 + b^2 c^2 - b^2 a^2 + c^2 a^2 - c^2 b^2 \right\}$

$= 0 = RHS$

Hence proved.

Properties of Triangles

In $\triangle ABC$, prove that $a \sin \left(B - C\right) + b \sin \left(C - A\right) + c \sin \left(A - B\right) = 0$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\therefore \;$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

$\begin{aligned} \therefore \; a \sin \left(B - C\right) & = 2 R \sin A \left(\sin B \cos C - \cos B \sin C\right) \\\\ & = 2 R \left(\sin A \sin B \cos C - \sin A \cos B \sin C\right) \;\;\; \cdots \; (1a) \end{aligned}$

$\begin{aligned} b \sin \left(C - A\right) & = 2 R \sin B \left(\sin C \cos A - \cos C \sin A\right) \\\\ & = 2 R \left(\sin B \sin C \cos A - \sin B \cos C \sin A\right) \;\;\; \cdots \; (1b) \end{aligned}$

$\begin{aligned} c \sin \left(A - B\right) & = 2 R \sin C \left(\sin A \cos B - \cos A \sin B\right) \\\\ & = 2 R \left(\sin C \sin A \cos B - \sin C \cos A \sin B\right) \;\;\; \cdots \; (1c) \end{aligned}$

Adding equations $(1a)$, $(1b)$ and $(1c)$ gives

$\begin{aligned} LHS & = 2 R \left(\sin A \sin B \cos C - \sin A \cos B \sin C \right. \\\\ & \hspace{2cm} \left. + \sin B \sin C \cos A - \sin B \cos C \sin A \right. \\\\ & \hspace{3cm} \left. + \sin C \sin A \cos B - \sin C \cos A \sin B \right) \\\\ & = 0 = RHS \end{aligned}$

Properties of Triangles

A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point $P$, she finds the distance to the eastern-most point of the pond to be $8 \; km$, while the distance to the western most point from $P$ to be $6 \; km$. If the angle between the two lines of sight is $60^\circ$, find the width of the pond.

In the figure,

eastern-most point of the pond $= E$

western-most point of the pond $= W$

width of the pond $= WE = p$

point of observation $= P$

distance to the eastern-most point of the pond from P $= PE = w = 8 \; km$

distance to the western-most point of the pond from P $= PW = e = 6 \; km$

$\angle WPE = 60^\circ$

Applying cosine rule to $\triangle WPE$ we have,

$\cos P = \dfrac{e^2 + w^2 - p^2}{2 e w}$

i.e. $\;$ $\cos \left(60^\circ\right) = \dfrac{6^2 + 8^2 - p^2}{2 \times 6 \times 8}$

i.e. $p^2 = 36 + 64 - 96 \cos 60^\circ$

i.e. $p^2 = 100 - \left(96 \times \dfrac{1}{2}\right) = 52$

$\therefore \;$ $p = \sqrt{52} = 2 \sqrt{13}$

$\therefore \;$ Width of the pond $= 2 \sqrt{13} \; km$

Properties of Triangles

In a $\triangle ABC$, if $a = \sqrt{3} - 1$, $b = \sqrt{3} + 1$ and $C = 60^\circ$, find the other side and the remaining two angles.

Given: In $\triangle ABC$, $a = \sqrt{3} - 1$, $b = \sqrt{3} + 1$, $C = 60^\circ$

By cosine rule,

$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\begin{aligned} \implies c^2 & = a^2 + b^2 - 2 a b \cos C \\\\ & = \left(\sqrt{3} - 1\right)^2 + \left(\sqrt{3} + 1\right)^2 - 2 \left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right) \cos \left(60^\circ\right) \\\\ & = 3 + 1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3} - 4 \times \dfrac{1}{2} \\\\ & = 6 \\\\ \implies c & = \sqrt{6} \end{aligned}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\begin{aligned} \implies \sin B & = \dfrac{b \sin C}{c} \\\\ & = \dfrac{\left(\sqrt{3} + 1\right) \sin \left(60^\circ\right)}{\sqrt{6}} \\\\ & = \dfrac{\left(\sqrt{3} + 1\right)}{\sqrt{6}} \times \dfrac{\sqrt{3}}{2} \\\\ & = \dfrac{\sqrt{3} + 1}{2 \sqrt{2}} \end{aligned}$

$\implies$ $B = \sin^{-1} \left(\dfrac{\sqrt{3} + 1}{2 \sqrt{2}}\right) = 105^\circ$

In $\triangle ABC$, $A + B + C = 180^\circ$

$\therefore \;$ $A = 180^\circ - \left(B + C\right) = 180^\circ - \left(105^\circ + 60^\circ\right) = 15^\circ$

Properties of Triangles

Find the area of $\triangle ABC$ if $a = 18 \; cm$, $b = 24 \; cm$ and $c = 30 \; cm$

Area of $\triangle ABC = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ $\;$ where $s$ is the semi-perimeter of $\triangle ABC$

$s = \dfrac{a + b + c}{2} = \dfrac{18 + 24 + 30}{2} = 36 \; cm$

$\begin{aligned} \therefore \; \text{Area of } \triangle ABC & = \sqrt{36 \left(36 - 18\right) \left(36 - 24\right) \left(36 - 30\right)} \\\\ & = \sqrt{36 \times 18 \times 12 \times 6} \\\\ & = 6 \times 3 \times 2 \times 6 \\\\ & = 216 \; cm^2 \end{aligned}$

Properties of Triangles

In any $\triangle ABC$, prove that the area of the triangle $\Delta = \dfrac{b^2 + c^2 - a^2}{4 \cot A}$

$\begin{aligned} \text{Area of } \triangle ABC & = \dfrac{1}{2} b c \sin A \\\\ & = \dfrac{1}{2} b c \times \left(\dfrac{\sin A}{\cos A}\right) \times \cos A \\\\ & = \dfrac{1}{2} b c \tan A \cos A \\\\ & = \dfrac{b c \cos A}{2 \cot A} \\\\ & \left[\text{Cosine rule: } \cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}\right] \\\\ & = \dfrac{b c \times \left(\dfrac{b^2 + c^2 - a^2}{2 bc}\right)}{2 \cot A} \\\\ & = \dfrac{b^2 + c^2 - a^2}{4 \cot A} \end{aligned}$

Hence proved.

Properties of Triangles

If the sides of $\triangle ABC$ are $a = 4$, $b = 6$ and $c = 8$, then show that $4 \cos B + 3 \cos C = 2$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

Given: $\;$ $a = 4$, $b = 6$, $c = 8$

$\therefore \;$ $\cos B = \dfrac{8^2 + 4^2 - 6^2}{2 \times 8 \times 4} = \dfrac{64 + 16 - 36}{64} = \dfrac{11}{16}$

$\cos C = \dfrac{4^2 + 6^2 - 8^2}{2 \times 4 \times 6} = \dfrac{16 + 36 - 64}{48} = - \dfrac{1}{4}$

$\therefore \;$ $LHS = 4 \cos B + 3 \cos C = 4 \times \dfrac{11}{16} + 3 \times \left(- \dfrac{1}{4}\right) = \dfrac{11}{4} - \dfrac{3}{4} = 2 = RHS$

Hence proved.

Properties of Triangles

In $\triangle ABC$, $\;$ prove that $\;$ $\dfrac{a + b}{a - b} = \tan \left(\dfrac{A + B}{2}\right) \cot \left(\dfrac{A - B}{2}\right)$

In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $A + B = \pi - C$

$\therefore \;$ $\dfrac{A + B}{2} = \dfrac{\pi}{2} - \dfrac{C}{2}$

$\therefore \;$ $\tan \left(\dfrac{A + B}{2}\right) = \tan \left(\dfrac{\pi}{2} - \dfrac{C}{2}\right) = \cot \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \therefore \; RHS & = \tan \left(\dfrac{A + B}{2}\right) \cot \left(\dfrac{A - B}{2}\right) \\\\ & = \cot \left(\dfrac{C}{2}\right) \cot \left(\dfrac{A - B}{2}\right) \;\;\; \left[\text{by equation } (1)\right] \\\\ & = \dfrac{\cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A - B}{2}\right)}{\sin \left(\dfrac{C}{2}\right) \sin \left(\dfrac{A - B}{2}\right)} \\\\ & = \dfrac{\dfrac{1}{2} \left[\cos \left(\dfrac{C + A - B}{2}\right) + \cos \left(\dfrac{C - A + B}{2}\right)\right]}{\dfrac{1}{2} \left[\cos \left(\dfrac{C - A + B}{2}\right) - \cos \left(\dfrac{C + A - B}{2}\right)\right]} \\\\ & = \dfrac{\cos \left(\dfrac{C + A}{2} - \dfrac{B}{2}\right) + \cos \left(\dfrac{C + B}{2} - \dfrac{A}{2}\right)}{\cos \left(\dfrac{C + B}{2} - \dfrac{A}{2}\right) - \cos \left(\dfrac{C + A}{2} - \dfrac{B}{2}\right)} \;\;\; \cdots \; (2) \end{aligned}$

Now, in $\triangle ABC$,

$C + A = \pi - B$ $\implies$ $\dfrac{C + A}{2} = \dfrac{\pi}{2} - \dfrac{B}{2}$ $\;\;\; \cdots \; (3a)$

$C + B = \pi - A$ $\implies$ $\dfrac{C + B}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$ $\;\;\; \cdots \; (3b)$

$\therefore \;$ In view of equations $(3a)$ and $(3b)$, equation $(2)$ becomes

$\begin{aligned} RHS & = \dfrac{\cos \left(\dfrac{\pi}{2} - \dfrac{B}{2} - \dfrac{B}{2}\right) + \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2} - \dfrac{A}{2}\right)}{\cos \left(\dfrac{\pi}{2} - \dfrac{A}{2} - \dfrac{A}{2}\right) - \cos \left(\dfrac{\pi}{2} - \dfrac{B}{2} - \dfrac{B}{2}\right)} \\\\ & = \dfrac{\cos \left(\dfrac{\pi}{2} - B\right) + \cos \left(\dfrac{\pi}{2} - A\right)}{\cos \left(\dfrac{\pi}{2} - A\right) - \cos \left(\dfrac{\pi}{2} - B\right)} \\\\ & = \dfrac{\sin B + \sin A}{\sin A - \sin B} \;\;\; \cdots \; (4) \end{aligned}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\therefore \;$ $\sin A = \dfrac{a}{2 R}$ $\;\;\; \cdots \; (5a)$

and $\;$ $\sin B = \dfrac{b}{2 R}$ $\;\;\; \cdots \; (5b)$

$\therefore \;$ In view of equations $(5a)$ and $(5b)$, equation $(4)$ becomes

$RHS = \dfrac{\dfrac{b}{2R} + \dfrac{a}{2R}}{\dfrac{a}{2R} - \dfrac{b}{2R}} = \dfrac{a + b}{a - b} = LHS $

Hence proved.

Properties of Triangles

In $\triangle ABC$, $\;$ prove that $\;$ $\left(a^2 - b^2 + c^2\right) \tan B = \left(a^2 + b^2 - c^2\right) \tan C$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$ $\implies$ $b^2 + c^2 - a^2 = 2 b c \cos A$

$\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$ $\implies$ $c^2 + a^2 - b^2 = 2 c a \cos B$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$ $\implies$ $a^2 + b^2 - c^2 = 2 a b \cos C$

Now,

$\begin{aligned} \left(a^2 - b^2 + c^2\right) \tan B & = 2 c a \cos B \tan B \\\\ & = 2 c a \sin B \\\\ & = \dfrac{a b c}{R} \;\;\; \cdots \; (1a) \end{aligned}$

$\begin{aligned} \left(a^2 + b^2 - c^2\right) \tan C & = 2 a b \cos C \tan C \\\\ & = 2 a b \sin C \\\\ & = \dfrac{a b c}{R} \;\;\; \cdots \; (1b) \end{aligned}$

$\therefore \;$ In view of equations $(1a)$ and $(1b)$ we have,

$\left(a^2 - b^2 + c^2\right) \tan B = \left(a^2 + b^2 - c^2\right) \tan C$

Hence proved.

Properties of Triangles

In $\triangle ABC$, $\;$ prove that $\;$ $\dfrac{a \sin \left(B - C\right)}{b^2 - c^2} = \dfrac{b \sin \left(C - A\right)}{c^2 - a^2} = \dfrac{c \sin \left(A - B\right)}{a^2 - b^2}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

By cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

Now,

$\begin{aligned} \sin \left(B - C\right) & = \sin B \cos C - \cos B \sin C \\\\ & = \dfrac{b}{2 R} \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) - \dfrac{c}{2 R} \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) \\\\ & = \dfrac{1}{4 a R} \left(a^2 + b^2 - c^2 - c^2 - a^2 + b^2\right) \\\\ & = \dfrac{2 \left(b^2 - c^2\right)}{4 a R} \\\\ & = \dfrac{b^2 - c^2}{2 a R} \;\;\; \cdots \; (1a) \end{aligned}$

$\begin{aligned} \sin \left(C - A\right) & = \sin C \cos A - \cos C \sin A \\\\ & = \dfrac{c}{2 R} \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) - \dfrac{a}{2 R} \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) \\\\ & = \dfrac{1}{4 b R} \left(b^2 + c^2 - a^2 - a^2 - b^2 + c^2\right) \\\\ & = \dfrac{2 \left(c^2 - a^2\right)}{4 b R} \\\\ & = \dfrac{c^2 - a^2}{2 b R} \;\;\; \cdots \; (1b) \end{aligned}$

$\begin{aligned} \sin \left(A - B\right) & = \sin A \cos B - \cos A \sin B \\\\ & = \dfrac{a}{2 R} \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) - \dfrac{b}{2 R} \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) \\\\ & = \dfrac{1}{4 c R} \left(c^2 + a^2 - b^2 - b^2 - c^2 + a^2\right) \\\\ & = \dfrac{2 \left(a^2 - b^2\right)}{4 c R} \\\\ & = \dfrac{a^2 - b^2}{2 c R} \;\;\; \cdots \; (1c) \end{aligned}$

$\therefore \;$ In view of equation $(1a)$,

$\dfrac{a \sin \left(B - C\right)}{b^2 - c^2} = \left(\dfrac{a}{b^2 - c^2}\right) \times \left(\dfrac{b^2 - c^2}{2 a R}\right) = \dfrac{1}{2 R}$ $\;\;\; \cdots \; (2a)$

In view of equation $(1b)$,

$\dfrac{b \sin \left(C - A\right)}{c^2 - a^2} = \left(\dfrac{b}{c^2 - a^2}\right) \times \left(\dfrac{c^2 - a^2}{2 b R}\right) = \dfrac{1}{2 R}$ $\;\;\; \cdots \; (2b)$

In view of equation $(1c)$,

$\dfrac{c \sin \left(A - B\right)}{a^2 - b^2} = \left(\dfrac{c}{a^2 - b^2}\right) \times \left(\dfrac{a^2 - b^2}{2 c R}\right) = \dfrac{1}{2 R}$ $\;\;\; \cdots \; (2c)$

$\therefore \;$ From equations $(2a)$, $(2b)$ and $(2c)$ we have,

$\dfrac{a \sin \left(B - C\right)}{b^2 - c^2} = \dfrac{b \sin \left(C - A\right)}{c^2 - a^2} = \dfrac{c \sin \left(A - B\right)}{a^2 - b^2}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, $\;$ prove that $\;$ $a \left(\cos B + \cos C\right) = 2 \left(b + c\right) \sin^2 \left(\dfrac{A}{2}\right)$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

By projection formula, $\;$ $b = c \cos A + a \cos C$, $\;$ $c = a \cos B + b \cos A$

$LHS = a \left(\cos B + \cos C\right)$

$= 2 a \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$

$\because \;$ $A$, $B$, $C$ are angles of $\triangle ABC$, $\;$ $A + B + C = \pi$

$\therefore \;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$

$\therefore \;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin\left(\dfrac{A}{2}\right)$

$\therefore \;$ $LHS = 2 a \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$ $\;\;\; \cdots \; (1)$

Now, $\;$ $b + c = c \cos A + a \cos C + a \cos B + b \cos A$

i.e. $\;$ $b + c = \left(b + c\right) \cos A + a \left(\cos B + \cos C\right)$

i.e. $\;$ $\left(b + c\right) \left(1 - \cos A\right) = 2 a \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$

i.e. $\;$ $2 \left(b + c\right) \sin^2 \left(\dfrac{A}{2}\right) = 2 a \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$

i.e. $RHS = 2 a \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$ $\;\;\; \cdots \; (2)$

$\therefore \;$ From equations $(1)$ and $(2)$, $LHS = RHS$

Hence proved.

Properties of Triangles

In $\triangle ABC$, $\;$ $\angle A = 60^{\circ}$. Prove that $\;$ $b + c = 2 a \cos \left(\dfrac{B - C}{2}\right)$

By projection formula,

$a = b \cos C + c \cos B$, $\;$ $b = c \cos A + a \cos C$, $\;$ $c = a \cos B + b \cos A$

$LHS = b + c = c \cos A + a \cos C + a \cos B + b \cos A$ $\;\;\;$ [by projection formula]

i.e. $\;$ $b + c = \left(b + c\right) \cos A + a \left(\cos B + \cos C\right)$

i.e. $\;$ $b + c = \left(b + c\right) \cos \left(60^{\circ}\right) + 2 a \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$ $\;\;$ [Given: $\angle A = 60^{\circ}$]

i.e. $\;$ $b + c = \dfrac{b + c}{2} + 2 a \cos \left(60^{\circ}\right) \cos \left(\dfrac{B - C}{2}\right)$

[Note:

$\because \;$ $A$, $B$, $C$ are angles of $\triangle ABC$, $\;$ $A + B + C = 180^{\circ}$

$\therefore \;$ $B + C = 180^{\circ} - A = 180^{\circ} - 60^{\circ} = 120^{\circ}$

$\therefore \;$ $\dfrac{B + C}{2} = 60^{\circ}$]

i.e. $\;$ $\dfrac{b + c}{2} = 2 a \times \dfrac{1}{2} \times \cos \left(\dfrac{B - C}{2}\right)$

i.e. $\;$ $b + c = 2 a \cos \left(\dfrac{B - C}{2}\right)$

Hence proved.

Properties of Triangles

In $\triangle ABC$ prove that $\;$ $\dfrac{\sin B}{\sin C} = \dfrac{c - a \cos B}{b - a \cos C}$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\begin{aligned} RHS & = \dfrac{c - a \cos B}{b - a \cos C} \\\\ & = \dfrac{c - a \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right)}{b - a \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)} \\\\ & = \dfrac{\dfrac{c^2 + b^2 - a^2}{c}}{\dfrac{b^2 + c^2 - a^2}{b}} \\\\ & = \dfrac{b}{c} \\\\ & = \dfrac{\sin B}{\sin C} = LHS \end{aligned}$

Properties of Triangles

The angles of $\triangle ABC$ are in arithmetic progression. If $b : c = \sqrt{3} : \sqrt{2}$, find $\angle A$.

Let the angles of $\triangle ABC$ be $A, \; B, \; C$.

Given: $\;$ $\angle A, \; \angle B, \; \angle C$ $\;$ are in arithmetic progression.

$\implies$ $2 B = A + C$ $\;\;\; \cdots \; (1)$

$\because \;$ $A, \; B, \; C$ are the angles of $\triangle ABC$,

$A + B + C = \pi$

i.e. $\;$ $2 B + B = \pi$ $\;\;\;$ [By equation $(1)$]

$\implies$ $B = \dfrac{\pi}{3}$

By law of sines, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\therefore \;$ We have $\;$ $\dfrac{b}{c} = \dfrac{\sin B}{\sin C}$

Given: $\;$ $\dfrac{b}{c} = \dfrac{\sqrt{3}}{\sqrt{2}}$

$\therefore \;$ $\dfrac{\sin B}{\sin C} = \dfrac{\sqrt{3}}{\sqrt{2}}$ $\;\;\; \cdots \; (2)$

Substituting the value of $B$ in equation $(2)$, we have,

$\sin C = \dfrac{\sqrt{2}}{\sqrt{3}} \times \sin \left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{2} = \dfrac{1}{\sqrt{2}}$

$\implies$ $C = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$

$\therefore \;$ $A = \pi - \left(\dfrac{\pi}{3} + \dfrac{\pi}{4}\right) = \dfrac{5 \pi}{12} = 75^{\circ}$

Inverse Trigonometric Functions

Find $\;$ $x$ $\;$ if $\;$ $\sin^{-1} \left(1 - x\right) - 2 \sin^{-1} \left(x\right) = \dfrac{\pi}{2}$

Let $\;$ $\sin^{-1} \left(1 - x\right) = \alpha$, $\;$ $\sin^{-1} \left(x\right) = \beta$ $\;\;\; \cdots \; (1)$

Then, $\;$ $\sin \left(\alpha\right) = 1 - x$, $\;$ $\sin \left(\beta\right) = x$ $\;\;\; \cdots \; (2)$

Given: $\;$ $\sin^{-1} \left(1 - x\right) - 2 \sin^{-1} \left(x\right) = \dfrac{\pi}{2}$

i.e. $\;$ $\alpha - 2 \beta = \dfrac{\pi}{2}$ $\;$ [by equation $(1)$]

i.e. $\;$ $\alpha = \dfrac{\pi}{2} + 2 \beta$

$\therefore \;$ $\sin \left(\alpha\right) = \sin \left(\dfrac{\pi}{2} + 2 \beta\right)$

i.e. $\;$ $\sin \left(\alpha\right) = \cos \left(2 \beta\right)$

i.e. $\;$ $\sin \left(\alpha\right) = 1 - 2 \sin^2 \left(\beta\right)$

i.e. $\;$ $1 - x = 1 - 2 x^2$ $\;$ [by equation $(2)$]

i.e. $\;$ $2 x^2 - x = 0$

i.e. $\;$ $x \left(2 x - 1\right) = 0$

$\implies$ $x = 0$ $\;\;\;$ OR $\;\;\;$ $x = \dfrac{1}{2}$

Putting $\;$ $x = 0$ $\;$ in the given equation,

$\begin{aligned} LHS & = \sin^{-1} \left(1 - 0\right) - 2 \sin^{-1} \left(0\right) \\\\ & = \sin^{-1} \left(1\right) - 0 \\\\ & = \dfrac{\pi}{2} = RHS \end{aligned}$

Putting $\;$ $x = \dfrac{1}{2}$ $\;$ in the given equation,

$\begin{aligned} LHS & = \sin^{-1} \left(1 - \dfrac{1}{2}\right) - 2 \sin^{-1} \left(\dfrac{1}{2}\right) \\\\ & = \sin^{-1} \left(\dfrac{1}{2}\right) - 2 \sin^{-1} \left(\dfrac{1}{2}\right) \\\\ & = - \sin^{-1}\left(\dfrac{1}{2}\right) \\\\ & = - \dfrac{\pi}{6} \neq RHS \end{aligned}$

$\therefore \;$ The solution set is $\left\{0 \right\}$

Inverse Trigonometric Functions

Solve: $\;$ $\tan \left\{2 \tan^{-1} \left(\dfrac{1}{5}\right) - \dfrac{\pi}{4}\right\}$

$\left[\text{Note: } 2 \tan^{-1} x = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right), \;\;\; x^2 < 1 \right.$
$\left. \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right) \right]$

$\tan \left\{2 \tan^{-1} \left(\dfrac{1}{5}\right) - \dfrac{\pi}{4}\right\}$

$= \tan \left\{\tan^{-1} \left(\dfrac{\dfrac{2}{5}}{1 - \dfrac{1}{25}}\right) - \tan^{-1} \left(1\right) \right\}$

$= \tan \left\{\tan^{-1} \left(\dfrac{5}{12}\right) - \tan^{-1} \left(1\right) \right\}$

$= \tan \left\{\tan^{-1} \left(\dfrac{\dfrac{5}{12} - 1}{1 + \dfrac{5}{12} \times 1}\right) \right\}$

$= \tan \left\{\tan^{-1} \left(- \dfrac{7}{17}\right) \right\}$

$= \tan \left\{- \tan^{-1} \left(\dfrac{7}{17}\right) \right\}$

$= - \tan \left\{\tan^{-1} \left(\dfrac{7}{17}\right)\right\}$

$= - \dfrac{7}{17}$

Inverse Trigonometric Functions

If $\alpha = \cos^{-1} \left(\dfrac{4}{5}\right)$, $\;$ $\beta = \tan^{-1} \left(\dfrac{2}{3}\right)$, $\;$ $\alpha, \; \beta \; \in \; \left(0, \dfrac{\pi}{2}\right)$, $\;$ then show that $\;$ $\left(\alpha - \beta\right) = \tan^{-1} \left(\dfrac{1}{18}\right)$

Given: $\;$ $\alpha = \cos^{-1} \left(\dfrac{4}{5}\right)$, $\;$ $\beta = \tan^{-1} \left(\dfrac{2}{3}\right)$

$\implies$ $\cos \alpha = \dfrac{4}{5}$, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{16}{25}} = \dfrac{3}{5}$, $\;$ $\tan \alpha = \dfrac{3}{4}$

and, $\;$ $\tan \beta = \dfrac{2}{3}$

Now, $\;$ $\tan \left(\alpha - \beta\right) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \; \tan \beta}$

i.e. $\;$ $\tan \left(\alpha - \beta\right) = \dfrac{\dfrac{3}{4} - \dfrac{2}{3}}{1 + \dfrac{3}{4} \times \dfrac{2}{3}} = \dfrac{1}{18} $

i.e. $\;$ $\left(\alpha - \beta\right) = \tan^{-1} \left(\dfrac{1}{18}\right)$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cot^{-1} \left\{\dfrac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}\right\}$, $\;\;$ $0 < x < \dfrac{\pi}{2}$

$\cot^{-1} \left\{\dfrac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}\right\}$

$= \cot^{-1} \left\{\dfrac{\left(\sqrt{1 - \sin x} + \sqrt{1 + \sin x}\right)^2}{\left(\sqrt{1 - \sin x} - \sqrt{1 + \sin x}\right) \left(\sqrt{1 - \sin x} + \sqrt{1 + \sin x}\right)} \right\}$

$= \cot^{-1} \left\{\dfrac{1 - \sin x + 1 + \sin x + 2 \sqrt{\left(1 - \sin x\right) \left(1 + \sin x\right)}}{1 - \sin x - 1 - \sin x} \right\}$

$= \cot^{-1} \left\{\dfrac{2 + 2 \sqrt{1 - \sin^2 x}}{- 2 \sin x} \right\}$

$= \cot^{-1} \left\{\dfrac{2 + 2 \cos x}{- 2 \sin x} \right\}$

$= \cot^{-1} \left\{\dfrac{1 + \cos x}{- \sin x} \right\}$

$= \cot^{-1} \left\{\dfrac{2 \cos^2 \left(\dfrac{x}{2}\right)}{- 2 \sin \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2}\right)} \right\}$

$= \cot^{-1} \left\{- \cot \left(\dfrac{x}{2}\right) \right\}$

$= \pi - \cot^{-1} \left\{\cot \left(\dfrac{x}{2}\right) \right\}$

$= \pi - \dfrac{x}{2}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\sin^{-1} \left\{ \cos \left[\sin^{-1} \left(x\right)\right]\right\} + \cos^{-1} \left\{\sin \left[\cos^{-1} \left(x\right)\right] \right\}$

$\left[\text{Note: } \sin^{-1} \left(x\right) = \cos^{-1} \left(\sqrt{1 - x^2}\right), \;\;\; 0 < x < 1 \right.$
$\left. \cos^{-1} \left(x\right) = \sin^{-1} \left(\sqrt{1 - x^2}\right), \;\;\; 0 < x < 1 \right.$
$\left. \sin^{-1} \left(x\right) + \cos^{-1} \left(x\right) = \dfrac{\pi}{2}, \;\;\; \left|x\right| \leq 1 \right]$

$\sin^{-1} \left\{ \cos \left[\sin^{-1} \left(x\right)\right]\right\} + \cos^{-1} \left\{\sin \left[\cos^{-1} \left(x\right)\right] \right\}$

$= \sin^{-1} \left\{\cos \left[\cos^{-1} \left(\sqrt{1 - x^2}\right)\right] \right\} + \cos^{-1} \left\{\sin \left[\sin^{-1} \left(\sqrt{1 - x^2}\right)\right] \right\}$

$= \sin^{-1} \left\{\sqrt{1 - x^2} \right\} + \cos^{-1} \left\{\sqrt{1 - x^2} \right\}$

$= \dfrac{\pi}{2}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\tan \left\{\sin^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{5}{13}\right)\right\}$

$\left[\text{Note: } \sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \;\;\; 0 < x < 1 \right.$
$\left. \cos^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right), \;\;\; 0 < x < 1 \right.$
$\left. \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \pi + \tan^{-1} \left(\dfrac{x + y}{1 - x y}\right), \;\;\; x \times y > 1 \right]$

$\tan \left\{\sin^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{5}{13}\right)\right\}$

$= \tan \left\{\tan^{-1} \left(\dfrac{\dfrac{3}{5}}{\sqrt{1 - \dfrac{9}{25}}}\right) + \tan^{-1} \left(\dfrac{\sqrt{1 - \dfrac{25}{169}}}{\dfrac{5}{13}}\right) \right\} $

$ = \tan \left\{\tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{12}{5}\right) \right\} $

$ = \tan \left\{\pi + \tan^{-1} \left(\dfrac{\dfrac{3}{4} + \dfrac{12}{5}}{1 - \dfrac{3}{4} \times \dfrac{12}{5}}\right) \right\} $

$ = \tan \left\{\pi + \tan^{-1} \left(\dfrac{-63}{16}\right) \right\} $

$ = \tan \left\{\tan^{-1} \left(\dfrac{- 63}{16}\right) \right\} $

$ = - \dfrac{63}{16} $

Inverse Trigonometric Functions

Find the value of x if $\;$ $\cos \left\{2 \tan^{-1} \left(x\right)\right\} = \dfrac{1}{2}$

$\left[\text{Note: } 2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right), \;\;\; x^2 < 1 \right.$
$\left. \tan^{-1} \left(x\right) = \cos^{-1} \left(\dfrac{1}{\sqrt{1 + x^2}}\right), \;\;\; x > 0 \right]$

Given: $\;$ $\cos \left\{2 \tan^{-1} \left(x\right)\right\} = \dfrac{1}{2}$

i.e. $\;$ $\cos \left\{\tan^{-1} \left(\dfrac{2x}{1 - x^2}\right)\right\} = \dfrac{1}{2}$

i.e. $\;$ $\cos \left\{\cos^{-1} \left[\dfrac{1}{\sqrt{1 + \left(\dfrac{2 x}{1 - x^2}\right)^2 }}\right]\right\} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{1 - x^2}{\sqrt{1 + x^4 - 2 x^2 + 4 x^2}} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{1 - x^2}{\sqrt{1 + x^4 + 2 x^2}} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{1 - x^2}{1 + x^2} = \dfrac{1}{2}$

i.e. $\;$ $2 - 2 x^2 = 1 + x^2$

i.e. $\;$ $3 x^2 = 1$ $\implies$ $x = \pm \dfrac{1}{\sqrt{3}}$

Verification:

Putting $\;$ $x = \dfrac{1}{\sqrt{3}}$ $\;$ in the given equation,

$LHS = \cos \left\{2 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)\right\}$

$= \cos \left\{2 \times \dfrac{\pi}{6}\right\} = \cos \left\{\dfrac{\pi}{3}\right\} = \dfrac{1}{2} = RHS$

Putting $\;$ $x = \dfrac{-1}{\sqrt{3}}$ $\;$ in the given equation,

$LHS = \cos \left\{2 \tan^{-1} \left(\dfrac{-1}{\sqrt{3}}\right)\right\}$

$= \cos \left\{2 \times \left(\dfrac{- \pi}{6}\right)\right\} = \cos \left\{\dfrac{-\pi}{3}\right\} = \cos \left\{\dfrac{\pi}{3}\right\} = \dfrac{1}{2} = RHS$

$\therefore \;$ The solution set is $\left\{\pm \dfrac{1}{\sqrt{3}}\right\}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cot \left[\text{cosec}^{-1} \left(\dfrac{5}{3}\right) + \tan^{-1} \left(\dfrac{2}{3}\right)\right]$

$\left[\text{Note: } \text{cosec}^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{1}{x}\right) \right.$
$\left. \sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \;\;\; 0 < x < 1 \right.$
$\left. \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x y}\right), \;\;\; x y < 1 \right.$

$\left. \tan^{-1} \left(x\right) = \cot^{-1} \left(\dfrac{1}{x}\right) \right]$

$\begin{aligned} \cot \left[\text{cosec}^{-1} \left(\dfrac{5}{3}\right) + \tan^{-1} \left(\dfrac{2}{3}\right)\right] & = \cot \left[\sin^{-1} \left(\dfrac{3}{5}\right) + \tan^{-1} \left(\dfrac{2}{3}\right)\right] \\\\ & = \cot \left[\tan^{-1} \left(\dfrac{\dfrac{3}{5}}{\sqrt{1 - \dfrac{9}{25}}}\right) + \tan^{-1} \left(\dfrac{2}{3}\right)\right] \\\\ & = \cot \left[\tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{2}{3}\right)\right] \\\\ & = \cot \left[\tan^{-1} \left(\dfrac{\dfrac{3}{4} + \dfrac{2}{3}}{1 - \dfrac{3}{4} \times \dfrac{2}{3}}\right)\right] \\\\ & = \cot \left[\tan^{-1} \left(\dfrac{17}{6}\right)\right] \\\\ & = \cot \left[\cot^{-1} \left(\dfrac{6}{17}\right)\right] \\\\ & = \dfrac{6}{17} \end{aligned}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cos \left[2 \cos^{-1} \left(x\right) + \sin^{-1} \left(x\right)\right]$ $\;$ when $\;$ $x = \dfrac{1}{3}$

$\left[\text{Note: } \sin^{-1} \left(x\right) = \cos^{-1} \left(\sqrt{1 - x^2}\right) \right.$

$\left. 2 \cos^{-1} \left(x\right) = \cos^{-1} \left(2 x^2 - 1\right) \right.$

$\left. \cos^{-1} \left(x\right) + \cos^{-1} \left(y\right) = \cos^{-1} \left(x y - \sqrt{1 - x^2} \sqrt{1 - y^2}\right), \;\; -1 \leq x, \; y \leq 1, \; x + y \geq 0 \right.$

$\left. \cos^{-1} \left(- x\right) = \pi - \cos^{-1} \left(x\right) \right]$

When $\;$ $x = \dfrac{1}{3}$,

$\begin{aligned} \cos \left[2 \cos^{-1} \left(x\right) + \sin^{-1} \left(x\right)\right] & = \cos \left[2 \cos^{-1} \left(\dfrac{1}{3}\right) + \sin^{-1} \left(\dfrac{1}{3}\right)\right] \\\\ & = \cos \left[2 \cos^{-1} \left(\dfrac{1}{3}\right) + \cos^{-1} \left(\sqrt{1 - \dfrac{1}{9}}\right)\right] \\\\ & = \cos \left[2 \cos^{-1} \left(\dfrac{1}{3}\right) + \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{2}{9} - 1\right) + \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(- \dfrac{7}{9}\right) + \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{-7 \sqrt{8}}{27} - \sqrt{1 - \dfrac{49}{81}} \times \sqrt{1 - \dfrac{8}{9}}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{- 7 \sqrt{8}}{27} - \dfrac{\sqrt{32}}{9} \times \dfrac{1}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{- 7 \sqrt{8} - 2 \sqrt{8}}{27}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{-9 \sqrt{8}}{27}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{- \sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\pi - \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = - \cos \left[\cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = - \dfrac{\sqrt{8}}{3} \end{aligned}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cot \left[\dfrac{\pi}{4} - 2 \cot^{-1} \left(3\right)\right]$

$\begin{aligned} \cot \left[\dfrac{\pi}{4} - 2 \cot^{-1} \left(3\right)\right] & = \cot \left[\dfrac{\pi}{4} - 2 \tan^{-1} \left(\dfrac{1}{3}\right)\right] \\\\ & \left[\text{Note: } 2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right), \;\; \text{if } \; x^2 < 1\right] \\\\ & = \cot \left[\dfrac{\pi}{4} - \tan^{-1} \left(\dfrac{\dfrac{2}{3}}{1 - \dfrac{1}{9}}\right)\right] \\\\ & = \cot \left[\dfrac{\pi}{4} - \tan^{-1} \left(\dfrac{3}{4}\right)\right] \\\\ & = \cot \left[\dfrac{\pi}{4} - \cot^{-1} \left(\dfrac{4}{3}\right)\right] \\\\ & \left[\text{Note: } \cot \left(\alpha - \beta\right) = \dfrac{\cot \alpha \times \cot \beta + 1}{\cot \beta - \cot \alpha}\right] \\\\ & = \dfrac{\cot \left(\dfrac{\pi}{4}\right) \times \cot \left[\cot^{-1} \left(\dfrac{4}{3}\right)\right] + 1}{\cot \left[\cot^{-1} \left(\dfrac{4}{3}\right)\right] - \cot \left(\dfrac{\pi}{4}\right)} \\\\ & = \dfrac{\dfrac{4}{3} + 1}{\dfrac{4}{3} - 1} \\\\ & = 7 \end{aligned}$

Inverse Trigonometric Functions

Solve: $\;$ $\tan^{-1} \left(2 x\right) + \tan^{-1} \left(\dfrac{1}{x + 4}\right) = \dfrac{\pi}{2}$

$\tan^{-1} x + \tan^{-1} y = \dfrac{\pi}{2} \iff x y = 1$

$\therefore \;$ $\tan^{-1} \left(2 x\right) + \tan^{-1} \left(\dfrac{1}{x + 4}\right) = \dfrac{\pi}{2}$ $\iff$ $2 x \times \dfrac{1}{x + 4} = 1$

i.e. $\;$ $2 x = x + 4$ $\implies$ $x = 4$

Verification:

Putting $\;$ $x = 4$ $\;$ in the given equation,

$LHS = \tan^{-1} \left(8\right) + \tan^{-1} \left(\dfrac{1}{4 + 4}\right)$

$= \tan^{-1} \left(8\right) + \tan^{-1} \left(\dfrac{1}{8}\right) = \dfrac{\pi}{2} = RHS$ $\;\;\;$ $\because \; 8 \times \dfrac{1}{8} = 1$

$\therefore \;$ The solution set is $\left\{4\right\}$

Inverse Trigonometric Functions

Solve: $\;$ $\sin^{-1} \left(x\right) + \cos^{-1} \left(2x\right) = \dfrac{\pi}{6}$

Let $\;$ $\sin^{-1} \left(x\right) = \alpha$, $\;\;\;$ $\alpha \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\;$ $x = \sin \alpha$ $\implies$ $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - x^2}$

Let $\;$ $\cos^{-1} \left(2x\right) = \beta$, $\;\;\;$ $\beta \in \left[0, \pi\right]$

Then, $\;$ $2 x = \cos \beta$ $\implies$ $\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - 4 x^2}$

Given: $\;$ $\sin^{-1} \left(x\right) + \cos^{-1} \left(2x\right) = \dfrac{\pi}{6}$

i.e. $\;$ $\alpha + \beta = \dfrac{\pi}{6}$

$\therefore \;$ $\sin \left(\alpha + \beta\right) = \sin \left(\dfrac{\pi}{6}\right)$

i.e. $\;$ $\sin \alpha \cos \beta + \cos \alpha \sin \beta = \dfrac{1}{2}$

i.e. $\;$ $ x \times 2x + \sqrt{1 - x^2} \times \sqrt{1 - 4 x^2} = \dfrac{1}{2}$

i.e. $\;$ $\sqrt{1 - 5 x^2 + 4 x^4} = \dfrac{1}{2} - 2 x^2$

i.e. $\;$ $1 - 5 x^2 + 4 x^4 = \left(\dfrac{1}{2} - 2 x^2\right)^2$

i.e. $\;$ $1 - 5 x^2 + 4 x^4 = \dfrac{1}{4} - 2 x^2 + 4 x^4$

i.e. $\;$ $3 x^2 = \dfrac{3}{4}$

i.e. $\;$ $x^2 = \dfrac{1}{4}$ $\implies$ $x = \pm \dfrac{1}{2}$

Verification:

Putting $\;$ $x = \dfrac{1}{2}$ $\;$ in the given equation,

$LHS = \sin^{-1} \left(\dfrac{1}{2}\right) + \cos^{-1} \left(2 \times \dfrac{1}{2}\right) = \dfrac{\pi}{6} + 0 = \dfrac{\pi}{6} = RHS$

Putting $\;$ $x = - \dfrac{1}{2}$ $\;$ in the given equation,

$LHS = \sin^{-1} \left(-\dfrac{1}{2}\right) + \cos^{-1} \left[2 \times \left(-\dfrac{1}{2}\right)\right] = -\dfrac{\pi}{6} + \pi = \dfrac{5 \pi}{6} \neq RHS$

$\therefore \;$ The solution set is $\left\{\dfrac{1}{2} \right\}$

Inverse Trigonometric Functions

Solve: $\;$ $2 \tan^{-1} \left(\cos x\right) = \tan^{-1} \left(2 \text{ cosec }x\right)$

$2 \tan^{-1} \left(\cos x\right) = \tan^{-1} \left(2 \text{ cosec }x\right)$

i.e. $\;$ $\tan^{-1} \left(\dfrac{2 \cos x}{1 - \cos^2 x}\right) = \tan^{-1} \left(2 \text{ cosec } x\right)$ $\;\;\;$ $\left[\text{Note: } 2 \tan^{-1} x = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right)\right]$

i.e. $\;$ $\dfrac{2 \cos x}{1 - \cos^2 x} = 2 \text{ cosec }x$ $\;\;\;$ $\left[\text{Note: } \tan^{-1} \text{ is one - one}\right]$

i.e. $\;$ $\dfrac{\cos x}{\sin^2 x} = \dfrac{1}{\sin x}$

i.e. $\;$ $\dfrac{\cos x}{\sin x} = 1$ $\;\;\;$ $\left[\text{provided } \sin x \neq 0\right]$

i.e. $\;$ $\tan x = 1$

i.e. $\;$ $x = \tan^{-1} \left(1\right) = \dfrac{\pi}{4}$

Verification:

Putting $x = \dfrac{\pi}{4}$ in the given equation,

$LHS = 2 \tan^{-1} \left[\cos \left(\dfrac{\pi}{4}\right)\right]$

$ = 2 \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \tan^{-1} \left[\dfrac{2 \times \dfrac{1}{\sqrt{2}}}{1 - \dfrac{1}{2}}\right] = \tan^{-1} \left(2 \sqrt{2}\right)$

$RHS = \tan^{-1} \left[2 \text{ cosec } \left(\dfrac{\pi}{4}\right)\right] = \tan^{-1} \left(2 \sqrt{2}\right) = LHS$

$\therefore \;$ The solution set is $\left\{\dfrac{\pi}{4} \right\}$

Inverse Trigonometric Functions

Solve: $\;$ $\tan^{-1} \left(\dfrac{x - 1}{x - 2}\right) + \tan^{-1} \left(\dfrac{x + 1}{x + 2}\right) = \dfrac{\pi}{4}$

$\tan^{-1} \left(\dfrac{x - 1}{x - 2}\right) + \tan^{-1} \left(\dfrac{x + 1}{x + 2}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{\dfrac{x - 1}{x - 2} + \dfrac{x + 1}{x + 2}}{1 - \left(\dfrac{x - 1}{x - 2}\right) \times \left(\dfrac{x + 1}{x + 2}\right)}\right] = \dfrac{\pi}{4}$ $\;\;\;$ provided $\left(\dfrac{x - 1}{x - 2}\right) \times \left(\dfrac{x + 1}{x + 2}\right) < 1$

i.e. $\;$ $\tan^{-1} \dfrac{\left(x - 1\right) \left(x + 2\right) + \left(x - 2\right) \left(x + 1\right)}{x^2 - 4 - x^2 + 1} = \dfrac{\pi}{4}$

i.e. $\;$ $\dfrac{x^2 + x - 2 + x^2 - x - 2}{- 3} = \tan \left(\dfrac{\pi}{4}\right)$

i.e. $\;$ $\dfrac{2 x^2 - 4}{- 3} = 1$

i.e. $\;$ $2 x^2 = 1$ $\implies$ $x^2 = \dfrac{1}{2}$ $\implies$ $x = \pm \dfrac{1}{\sqrt{2}}$

Verification:

Putting $x = + \dfrac{1}{\sqrt{2}}$ in the given equation,

$LHS = \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}} - 2} \right) + \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}} + 1}{\dfrac{1}{\sqrt{2}} + 2} \right)$

$= \tan^{-1} \left(\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}}\right) + \tan^{-1} \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right)$

$= \tan^{-1} \left[\dfrac{\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}} + \dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}}{1 - \left(\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}}\right) \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right)}\right]$ $\;\;\;$ $\because \; \left(\dfrac{1 - \sqrt{2}}{1 - 2 \sqrt{2}}\right) \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right) < 1$

$= \tan^{-1} \left(\dfrac{1 + 2 \sqrt{2} - \sqrt{2} - 4 + 1 - 2 \sqrt{2} + \sqrt{2} - 4}{1 - 8 - 1 + 2}\right)$

$= \tan^{-1} \left(\dfrac{- 6}{- 6}\right) = \tan^{-1} \left(1\right) = \dfrac{\pi}{4} = RHS$

Putting $x = - \dfrac{1}{\sqrt{2}}$ in the given equation,

$LHS = \tan^{-1} \left(\dfrac{- \dfrac{1}{\sqrt{2}} - 1}{- \dfrac{1}{\sqrt{2}} - 2} \right) + \tan^{-1} \left(\dfrac{- \dfrac{1}{\sqrt{2}} + 1}{- \dfrac{1}{\sqrt{2}} + 2} \right)$

$= \tan^{-1} \left(\dfrac{- 1 - \sqrt{2}}{- 1 - 2 \sqrt{2}}\right) + \tan^{-1} \left(\dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}\right)$

$= \tan^{-1} \left[\dfrac{\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}} + \dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}}{1 - \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right) \left(\dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}\right)}\right]$ $\;\;\;$ $\because \; \left(\dfrac{1 + \sqrt{2}}{1 + 2 \sqrt{2}}\right) \left(\dfrac{- 1 + \sqrt{2}}{- 1 + 2 \sqrt{2}}\right) < 1$

$= \tan^{-1} \left(\dfrac{- 1 + 2 \sqrt{2} - \sqrt{2} + 4 - 1 - 2 \sqrt{2} + \sqrt{2} + 4}{8 - 1 - 2 + 1}\right)$

$= \tan^{-1} \left(\dfrac{6}{6}\right) = \tan^{-1} \left(1\right) = \dfrac{\pi}{4} = RHS$

$\therefore \;$ The solution set is $\left\{\pm \dfrac{1}{\sqrt{2}} \right\}$

Inverse Trigonometric Functions

Prove that: $\;$ $\tan \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] + \tan \left[\dfrac{\pi}{4} - \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] = \dfrac{2b}{a}$

Let $\;$ $\cos^{-1} \left(\dfrac{a}{b}\right) = \theta$, $\;\;\;$ $\theta \in \left[0, \pi\right]$

Then, $\;$ $\cos \theta = \dfrac{a}{b}$

$\begin{aligned} \text{Also, } \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] & = \tan \left[\dfrac{\theta}{2}\right] \\\\ & = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}} \\\\ & = \sqrt{\dfrac{1 - \dfrac{a}{b}}{1 + \dfrac{a}{b}}} \\\\ & = \sqrt{\dfrac{b - a}{b + a}} \\\\ & = \dfrac{\sqrt{b^2 - a^2}}{b + a} \;\;\; \cdots \; (1) \end{aligned}$

LHS $= \tan \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] + \tan \left[\dfrac{\pi}{4} - \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]$

$= \dfrac{\tan \left(\dfrac{\pi}{4}\right) + \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]}{1 - \tan \left(\dfrac{\pi}{4}\right) \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]} + \dfrac{\tan \left(\dfrac{\pi}{4}\right) - \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]}{1 + \tan \left(\dfrac{\pi}{4}\right) \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]}$

$= \dfrac{1 + \dfrac{\sqrt{b^2 - a^2}}{b + a}}{1 - \dfrac{\sqrt{b^2 - a^2}}{b + a}} + \dfrac{1 - \dfrac{\sqrt{b^2 - a^2}}{b + a}}{1 + \dfrac{\sqrt{b^2 - a^2}}{b + a}}$ $\;\;\; \left[\text{In view of equation } (1)\right]$

$= \dfrac{b + a + \sqrt{b^2 - a^2}}{b + a - \sqrt{b^2 - a^2}} + \dfrac{b + a - \sqrt{b^2 - a^2}}{b + a + \sqrt{b^2 - a^2}}$

$= \dfrac{\left(a + b + \sqrt{b^2 - a^2}\right)^2 + \left(a + b - \sqrt{b^2 - a^2}\right)^2}{\left(a + b + \sqrt{b^2 - a^2}\right) \left(a + b - \sqrt{b^2 - a^2}\right)}$

$= \dfrac{\left(a + b\right)^2 + b^2 - a^2 + 2 \left(a + b\right) \sqrt{b^2 - a^2} + \left(a + b\right)^2 + b^2 - a^2 - 2 \left(a + b\right) \sqrt{b^2 - a^2} }{\left(a + b\right)^2 - \left(b^2 - a^2\right)}$

$= \dfrac{a^2 + b^2 + 2 a b + b^2 - a^2 + a^2 + b^2 + 2 a b + b^2 - a^2}{a^2 + b^2 + 2 a b - b^2 + a^2}$

$= \dfrac{4 b^2 + 4 a b}{2 a^2 + 2 a b}$

$= \dfrac{4 b \left(a + b\right)}{2 a \left(a + b\right)}$

$= \dfrac{2 b}{a} = \text{RHS}$

Hence proved.

Inverse Trigonometric Functions

If $\;$ $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$, $\;$ then prove that $\;$ $x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2 x y z$

$\left\{ \text{Note: } \sin A + \sin B = 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) \right.$
$\left. \cos A - \cos B = 2 \sin \left(\dfrac{A + B}{2}\right) \sin \left(\dfrac{B - A}{2}\right) \right\}$

Let $\;$ $\sin^{-1} x = \alpha$, $\;$ $\sin^{-1} y = \beta$, $\;$ $\sin^{-1} z = \gamma$, $\;\;$ $\alpha, \; \beta, \; \gamma \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\;$ $\sin \alpha = x$, $\;$ $\sin \beta = y$, $\;$ $\sin \gamma = z$ $\;\;\; \cdots \; (1)$

and $\;$ $\cos \alpha = \sqrt{1 - x^2}$, $\;$ $\cos \beta = \sqrt{1 - y^2}$, $\;$ $\cos \gamma = \sqrt{1 - z^2}$ $\;\;\; \cdots \; (2)$

Given: $\;$ $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$

$\implies$ $\alpha + \beta + \gamma = \pi$ $\;\;\; \cdots \; (3)$

Now,

$\begin{aligned} \sin \left(2 \alpha\right) + \sin \left(2 \beta\right) + \sin \left(2 \gamma\right) & = 2 \sin \left(\dfrac{2 \alpha + 2 \beta}{2}\right) \cos \left(\dfrac{2 \alpha - 2 \beta}{2}\right) + 2 \sin \gamma \cos \gamma \\\\ & = 2 \sin \left(\alpha + \beta\right) \cos \left(\alpha - \beta\right) + 2 \sin \gamma \cos \gamma \\\\ & = 2 \sin \left(\pi - \gamma\right) \cos \left(\alpha - \beta\right) + 2 \sin \gamma \cos \gamma \\ & \hspace{2cm} \left[\text{In view of equation } (3)\right] \\\\ & = 2 \sin \gamma \cos \left(\alpha - \beta\right) + 2 \sin \gamma \cos \gamma \\\\ & = 2 \sin \gamma \left[\cos \left(\alpha - \beta\right) + \cos \gamma\right] \\\\ & = 2 \sin \gamma \left\{ \cos \left(\alpha - \beta\right) + \cos \left[\pi - \left(\alpha + \beta\right)\right] \right\} \\ & \hspace{2cm} \left[\text{In view of equation } (3)\right] \\\\ & = 2 \sin \gamma \left[\cos \left(\alpha - \beta\right) - \cos \left(\alpha + \beta\right)\right] \\\\ & = 2 \sin \gamma \times 2 \sin \left(\dfrac{\alpha - \beta + \alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha + \beta - \alpha + \beta}{2}\right) \\\\ & = 4 \sin \alpha \times \sin \beta \times \sin \gamma \end{aligned}$

i.e. $\;$ $2 \sin \alpha \cos \alpha + 2 \sin \beta \cos \beta + 2 \sin \gamma \cos \gamma = 4 \sin \alpha \times \sin \beta \times \sin \gamma$

i.e. $\;$ $\sin \alpha \cos \alpha + \sin \beta \cos \beta + \sin \gamma \cos \gamma = 2 \sin \alpha \times \sin \beta \times \sin \gamma$ $\;\;\; \cdots \; (4)$

$\therefore \;$ In view of equations $(1)$ and $(2)$, equation $(4)$ becomes,

$x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2 x y z$

Hence proved.

Inverse Trigonometric Functions

If $\;$ $\tan^{-1} \left(\sqrt{\dfrac{a \; r}{b \; c}}\right) + \tan^{-1} \left(\sqrt{\dfrac{b \; r}{c \; a}}\right) + \tan^{-1} \left(\sqrt{\dfrac{c \; r}{a \; b}}\right) = \pi$, $\;$ then prove that $\;$ $a + b + c = r$, $\;\;\;$ $a, \; b, \; c, \; r > 0$

Let $\;$ $\tan^{-1} \left(\sqrt{\dfrac{a \; r}{b \; c}}\right) = \alpha$, $\;$ $\tan^{-1} \left(\sqrt{\dfrac{b \; r}{c \; a}}\right) = \beta$, $\;$ $\tan^{-1} \left(\sqrt{\dfrac{c \; r}{a \; b}}\right) = \gamma$, $\;$ $\alpha, \; \beta, \; \gamma \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

Then, $\;$ $\tan \alpha = \sqrt{\dfrac{a \; r}{b \; c}}$, $\;$ $\tan \beta = \sqrt{\dfrac{b \; r}{c \; a}}$, $\;$ $\tan \gamma = \sqrt{\dfrac{c \; r}{a \; b}}$

$\therefore \;$ $\tan^{-1} \left(\sqrt{\dfrac{a \; r}{b \; c}}\right) + \tan^{-1} \left(\sqrt{\dfrac{b \; r}{c \; a}}\right) + \tan^{-1} \left(\sqrt{\dfrac{c \; r}{a \; b}}\right) = \pi$

$\implies$ $\alpha + \beta + \gamma = \pi$

i.e. $\;$ $\alpha + \beta = \pi - \gamma$

$\therefore \;$ $\tan \left(\alpha + \beta\right) = \tan \left(\pi - \gamma\right)$

i.e. $\;$ $\dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} = - \tan \gamma$

i.e. $\;$ $\dfrac{\sqrt{\dfrac{a \; r}{b \; c}} + \sqrt{\dfrac{b \; r}{c \; a}} }{1 - \sqrt{\dfrac{a \; r}{b \; c}} \times \sqrt{\dfrac{b \; r}{c \; a}}} = - \sqrt{\dfrac{c \; r}{a \; b}} $

i.e. $\;$ $\dfrac{\dfrac{a \sqrt{c \; r} + b \sqrt{c \; r}}{c \sqrt{a \; b}}}{\dfrac{c \sqrt{a \; b} - r \sqrt{a \; b}}{c \sqrt{a \; b}}} = - \sqrt{\dfrac{c \; r}{a \; b}} $

i.e. $\;$ $\dfrac{\left(a + b\right) \sqrt{c \; r}}{\left(c - r\right) \sqrt{a \; b}} = - \sqrt{\dfrac{c \; r}{a \; b}} $

i.e. $\;$ $\dfrac{a + b}{c - r} = - 1$

i.e. $\;$ $a + b = r - c$

i.e. $\;$ $a + b + c = r$

Hence proved.

Inverse Trigonometric Functions

If $\;$ $a > b > c > 0$, $\;$ then prove that $\;$ $\cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \cot^{-1} \left[\dfrac{c \cdot a + 1}{c - a}\right] = \pi$

Given: $\;$ $a > b \implies a - b > 0$; $\;$ $b > c \implies b - c > 0$; $\;$ $a > c \implies c - a < 0$

$\because$ $\;$ $a > 0, \; b > 0, \; c > 0$ $\;$ $\implies$ $a \cdot b > 0$, $\;$ $b \cdot c > 0$, $\;$ $c \cdot a > 0$

$\therefore \;$ $\cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \cot^{-1} \left[\dfrac{c \cdot a + 1}{c - a}\right]$

$= \cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \cot^{-1} \left[- \dfrac{c \cdot a + 1}{a - c}\right]$

$\left\{\text{Note: } \cot^{-1} \left(- x\right) = \pi - \cot^{-1} \left(x\right), \;\;\; x \in R \right\}$

$= \cot^{-1} \left[\dfrac{a \cdot b + 1}{a - b}\right] + \cot^{-1} \left[\dfrac{b \cdot c + 1}{b - c}\right] + \pi - \cot^{-1} \left[\dfrac{c \cdot a + 1}{a - c}\right]$

$\left\{\text{Note: } \cot^{-1} \left(x\right) - \cot^{-1} \left(y\right) = \cot^{-1} \left(\dfrac{x \cdot y + 1}{y - x}\right)\right\}$

$= \cot^{-1} \left(b\right) - \cot^{-1} \left(a\right) + \cot^{-1} \left(c\right) - \cot^{-1} \left(b\right) + \pi - \cot^{-1} \left(c\right) + \cot^{-1} \left(a\right)$

$= \pi$

Hence proved.

Inverse Trigonometric Functions

If $\;$ $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, $\;$ then prove that $\;$ $x + y + z = x \cdot y \cdot z$

Let $\;$ $\tan^{-1} x = \alpha$, $\;$ $\tan^{-1} y = \beta$, $\;$ $\tan^{-1} z = \gamma$, $\;\;$ $\alpha, \; \beta, \; \gamma \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

Then, $\;$ $x = \tan \alpha$, $\;$ $y = \tan \beta$, $\;$ $z = \tan \gamma$

Given: $\;$ $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$

i.e. $\;$ $\alpha + \beta + \gamma = \pi$

i.e. $\;$ $\alpha + \beta = \pi - \gamma$

i.e. $\;$ $\tan \left[\alpha + \beta\right] = \tan \left[\pi - \gamma\right]$

i.e. $\;$ $\dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \times \tan \beta} = - \tan \gamma$

i.e. $\;$ $\dfrac{x + y}{1 - x \cdot y} = - z$

i.e. $\;$ $x + y = - z + x \cdot y \cdot z$

i.e. $\;$ $x + y + z = x \cdot y \cdot z$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\sin^{-1} \left[\dfrac{\sin x + \cos x}{\sqrt{2}}\right] = \dfrac{\pi}{4} + x$, $\;\;\;$ $- \dfrac{\pi}{4} < x < \dfrac{\pi}{4}$

$\begin{aligned} \sin^{-1} \left[\dfrac{\sin x + \cos x}{\sqrt{2}}\right] & = \sin^{-1} \left[\dfrac{1}{\sqrt{2}} \times \sin x + \dfrac{1}{\sqrt{2}} \times \cos x\right] \\\\ & = \sin^{-1} \left[\cos \left(\dfrac{\pi}{4}\right) \sin x + \sin \left(\dfrac{\pi}{4}\right) \cos x\right] \\\\ & = \sin^{-1} \left[\sin \left(\dfrac{\pi}{4} + x\right)\right] \\\\ & = \dfrac{\pi}{4} + x \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that:
$\;$ $\tan^{-1} \left[\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right] = \dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(x^2\right)$, $\;$ $- 1 < x < 1$, $\;$ $x \neq 0$

$\tan^{-1} \left[\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right]$

$= \tan^{-1} \left[\dfrac{\left(\sqrt{1 + x^2} + \sqrt{1 - x^2}\right) \left(\sqrt{1 + x^2} - \sqrt{1 - x^2}\right)}{\left(\sqrt{1 + x^2} - \sqrt{1 - x^2}\right)^2}\right]$

$= \tan^{-1} \left[\dfrac{1 + x^2 - 1 + x^2}{1 + x^2 + 1 - x^2 - 2 \sqrt{\left(1 + x^2\right) \left(1 - x^2\right)}}\right]$

$= \tan^{-1} \left[\dfrac{2 x^2}{2 - 2 \sqrt{1 - x^4}}\right]$

$= \tan^{-1} \left[\dfrac{x^2}{1 - \sqrt{1 - x^4}}\right]$ $\;\;\; \cdots \; (1)$

Let $\;$ $x^2 = \cos \theta \implies \theta = \cos^{-1} \left(x^2\right), \hspace{1cm} \theta \in \left[0, \pi\right]$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes,

$\tan^{-1} \left[\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right]$

$= \tan^{-1} \left[\dfrac{\cos \theta}{1 - \sqrt{1 - \cos^2 \theta}}\right]$

$= \tan^{-1} \left[\dfrac{\cos \theta}{1 - \sin \theta}\right]$

$= \tan^{-1} \left[\dfrac{\cos^2 \left(\dfrac{\theta}{2}\right) - \sin^2 \left(\dfrac{\theta}{2}\right)}{\cos^2 \left(\dfrac{\theta}{2}\right) + \sin^2 \left(\dfrac{\theta}{2}\right) - 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)}\right]$

$= \tan^{-1} \left[\dfrac{\left\{\cos \left(\dfrac{\theta}{2}\right) + \sin \left(\dfrac{\theta}{2}\right) \right\} \left\{\cos \left(\dfrac{\theta}{2}\right) - \sin \left(\dfrac{\theta}{2}\right) \right\}}{\left\{\cos\left(\dfrac{\theta}{2}\right) - \sin \left(\dfrac{\theta}{2}\right) \right\}^2}\right]$

$= \tan^{-1} \left[\dfrac{\cos \left(\dfrac{\theta}{2}\right) + \sin \left(\dfrac{\theta}{2}\right)}{\cos \left(\dfrac{\theta}{2}\right) - \sin \left(\dfrac{\theta}{2}\right)}\right]$

$= \tan^{-1} \left[\dfrac{1 + \tan \left(\dfrac{\theta}{2}\right)}{1 - \tan \left(\dfrac{\theta}{2}\right)}\right]$

$= \tan^{-1} \left[\tan \left(\dfrac{\pi}{4} + \dfrac{\theta}{2}\right)\right]$

$= \dfrac{\pi}{4} + \dfrac{\theta}{2}$

$= \dfrac{\pi}{2} + \dfrac{1}{2} \cos^{-1} \left(x^2\right)$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\cot^{-1} \left(\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right) = \dfrac{x}{2}$, $\;\;\;$ $0 < x < \dfrac{\pi}{2}$

$\cot^{-1} \left[\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right]$

$= \cot^{-1} \left[\dfrac{\left(\sqrt{1 + \sin x} + \sqrt{1 - \sin x}\right)^2}{\left(\sqrt{1 + \sin x} - \sqrt{1 - \sin x}\right) \left(\sqrt{1 + \sin x} + \sqrt{1 - \sin x}\right)}\right]$

$= \cot^{-1} \left[\dfrac{1 + \sin x + 1 - \sin x + 2 \sqrt{\left(1 + \sin x\right) \left(1 - \sin x\right)}}{1 + \sin x - 1 + \sin x}\right]$

$= \cot^{-1} \left[\dfrac{2 + 2 \cos x}{2 \sin x}\right]$

$= \cot^{-1} \left[\dfrac{1 + \cos x}{\sin x}\right]$

$= \cot^{-1} \left[\dfrac{2 \cos^2 \left(\dfrac{x}{2}\right)}{2 \sin \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2}\right)}\right]$

$= \cot^{-1} \left[\cot \left(\dfrac{x}{2}\right)\right]$

$= \dfrac{x}{2}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\cot^{-1} \left(\dfrac{\sqrt{1 + x^2} -1}{x}\right) = \dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} \left(x\right)$

Let $x = \tan \left(\theta\right) \implies \theta = \tan^{-1} \left(x\right), \hspace{1cm} \theta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\begin{aligned} \therefore \; \cot^{-1} \left(\dfrac{\sqrt{1 + x^2} -1}{x}\right) & = \cot^{-1} \left(\dfrac{\sqrt{1 + \tan^2 \theta} - 1}{\tan \theta}\right) \\\\ & = \cot^{-1} \left(\dfrac{\sec \theta - 1}{\tan \theta}\right) \\\\ & = \cot^{-1} \left(\dfrac{1 - \cos \theta}{\sin \theta}\right) \\\\ & = \cot^{-1} \left[\dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)}\right] \\\\ & = \cot^{-1} \left[\tan \left(\dfrac{\theta}{2}\right)\right] \\\\ & = \cot^{-1} \left[\cot \left(\dfrac{\pi}{2} - \dfrac{\theta}{2}\right)\right] \\\\ & = \dfrac{\pi}{2} - \dfrac{\theta}{2} \\\\ & = \dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} \left(x\right) \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\cos^{-1} \left(2 x^2 - 1\right) = 2 \cos^{-1} \left(x\right), \hspace{1cm} 0 < x <1$

Let $\;$ $x = \cos \theta \implies \theta = \cos^{-1} \left(x\right), \hspace{1cm} \theta \in \left[0, \pi\right]$

$\begin{aligned} \text{Then, } \cos^{-1} \left(2 x^2 - 1\right) & = \cos^{-1} \left(2 \cos^2 \theta - 1\right) \\\\ & = \cos^{-1} \left[\cos \left(2 \theta\right)\right] \hspace{1cm} \left[\text{Note: } \cos 2 \theta = 2 \cos^2 \theta - 1\right] \\\\ & = 2 \theta \\\\ & = 2 \cos^{-1} \left(x\right) \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\sin^{-1} \left(\dfrac{12}{13}\right) + \cos^{-1} \left(\dfrac{4}{5}\right) + \tan^{-1} \left(\dfrac{63}{16}\right) = \pi$

$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$\cos^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \pi + \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y > 1$

$\tan^{-1} \left(-x\right) = - \tan^{-1} \left(x\right), \hspace{1cm} x \in R$

Now, $\;$ $\sin^{-1} \left(\dfrac{12}{13}\right) + \cos^{-1} \left(\dfrac{4}{5}\right) + \tan^{-1} \left(\dfrac{63}{16}\right)$

$= \tan^{-1} \left[\dfrac{12/13}{\sqrt{1 - \left(12/13\right)^2}}\right] + \tan^{-1} \left[\dfrac{\sqrt{1 - \left(4/5\right)^2}}{4/5}\right] + \tan^{-1} \left(\dfrac{63}{16}\right)$

$\left[\text{Note: } 0 < \dfrac{12}{13} < 1; \hspace{0.5cm} 0 < \dfrac{4}{5} < 1\right]$

$= \tan^{-1} \left(\dfrac{12}{5}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{63}{16}\right)$

$\left[\text{Note: } \dfrac{12}{5} \times \dfrac{3}{4} > 1\right]$

$= \pi + \tan^{-1} \left(\dfrac{\dfrac{12}{5} + \dfrac{3}{4}}{1 - \dfrac{12}{5} \times \dfrac{3}{4}}\right) + \tan^{-1} \left(\dfrac{63}{16}\right)$

$= \pi + \tan^{-1} \left(-\dfrac{63}{16}\right) + \tan^{-1} \left(\dfrac{63}{16}\right)$

$= \pi - \tan^{-1} \left(\dfrac{63}{16}\right) + \tan^{-1} \left(\dfrac{63}{16}\right)$

$= \pi$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $2 \cot^{-1} \left(\dfrac{1}{3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) = \pi$

$\cot^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{1}{x}\right), \hspace{1cm} x > 0$

$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \pi + \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y > 1$

$\tan^{-1} \left(-x\right) = - \tan^{-1} \left(x\right), \hspace{1cm} x \in R$

$\begin{aligned} \therefore \; 2 \cot^{-1} \left(\dfrac{1}{3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) & = 2 \tan^{-1} \left(\dfrac{1}{1/3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & = 2 \tan^{-1} \left(3\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & \left[\because \; 3 \times 3 > 1\right] \\\\ & = \pi + \tan^{-1} \left(\dfrac{3 + 3}{1 - 3 \times 3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & = \pi + \tan^{-1} \left(- \dfrac{6}{8}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & = \pi - \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & = \pi \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $2 \cot^{-1} \left(2\right) + \text{cosec}^{-1} \left(\dfrac{5}{3}\right) = \dfrac{\pi}{2}$

$\cot^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{1}{x}\right), \hspace{1cm} x > 0$

$\text{cosec}^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{1}{x}\right), \hspace{1cm} \left|x\right| \geq 1$

$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right), \;\; x^2 < 1$

$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \dfrac{\pi}{2}, \;\; \text{if } x \cdot y = 1$

Now, $\;$ $\cot^{-1} \left(2\right) = \tan^{-1} \left(\dfrac{1}{2}\right), \;\;\; \left[2 > 0\right]$

$\text{cosec}^{-1} \left(\dfrac{5}{3}\right) = \sin^{-1} \left(\dfrac{1}{5/3}\right) = \sin^{-1} \left(\dfrac{3}{5}\right), \;\;\; \left|\dfrac{5}{3}\right| \geq 1$

$\begin{aligned} \therefore \; 2 \cot^{-1} \left(2\right) + \text{cosec}^{-1} \left(\dfrac{5}{3}\right) & = 2 \tan^{-1} \left(\dfrac{1}{2}\right) + \sin^{-1} \left(\dfrac{3}{5}\right) \\\\ & \left[\left(\dfrac{1}{2}\right)^2 < 1, \;\; 0 < \dfrac{3}{5} < 1\right] \\\\ & = \tan^{-1} \left[\dfrac{2 \times \dfrac{1}{2}}{1 - \left(\dfrac{1}{2}\right)^2}\right] + \tan^{-1} \left[\dfrac{3/5}{\sqrt{1 - \left(3/5\right)^2}}\right] \\\\ & = \tan^{-1} \left(\dfrac{1}{3/4}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & = \tan^{-1} \left(\dfrac{4}{3}\right) + \tan^{-1} \left(\dfrac{3}{4}\right) \\\\ & = \dfrac{\pi}{2} \hspace{1cm} \left[\because \dfrac{4}{3} \times \dfrac{3}{4} = 1\right] \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $2 \sin^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{24}{25}\right) = \dfrac{\pi}{2}$

$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$\cos^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$\begin{aligned} \therefore \; 2 \sin^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{24}{25}\right) & = 2 \tan^{-1} \left[\dfrac{\dfrac{3}{5}}{\sqrt{1 - \left(\dfrac{3}{5}\right)^2}}\right] + \tan^{-1} \left[\dfrac{\sqrt{1 - \left(\dfrac{24}{25}\right)^2}}{\dfrac{24}{25}}\right] \\\\ & \left(\text{Note: } 0 < \dfrac{3}{5} < 1, \;\;\; 0 < \dfrac{24}{25} < 1 \right) \\\\ & = 2 \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{7}{24}\right) \\\\ & \left[\text{Note: } 2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right), \;\; x^2 < 1\right] \\\\ & = \tan^{-1} \left[\dfrac{2 \times \dfrac{3}{4}}{1 - \left(\dfrac{3}{4}\right)^2}\right] + \tan^{-1} \left(\dfrac{7}{24}\right) \\\\ & = \tan^{-1} \left(\dfrac{24}{7}\right) + \tan^{-1} \left(\dfrac{7}{24}\right) \\\\ & \left[\text{Note: } \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \dfrac{\pi}{2}, \;\; \text{if } x \cdot y = 1\right] \\\\ & = \dfrac{\pi}{2} \;\;\; \left[\because \dfrac{24}{7} \times \dfrac{7}{24} = 1\right] \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\cos^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) = \tan^{-1} \left(\dfrac{56}{33}\right)$

$\cos^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$

$\begin{aligned} \therefore \; \cos^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) & = \tan^{-1} \left[\dfrac{\sqrt{1 - \left(\dfrac{4}{5}\right)^2}}{\dfrac{4}{5}}\right] + \tan^{-1} \left[\dfrac{\dfrac{5}{13}}{\sqrt{1 - \left(\dfrac{5}{13}\right)^2}}\right] \\\\ & \left(\text{Note: } 0 < \dfrac{4}{5} < 1, \;\;\; 0 < \dfrac{5}{13} < 1 \right) \\\\ & = \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{5}{12}\right) \\\\ & \left[\text{Note: } \right. \\ & \left. \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1\right] \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{3}{4} + \dfrac{5}{12}}{1 - \dfrac{3}{4} \times \dfrac{5}{12}}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{14}{12}}{\dfrac{33}{48}}\right) \\\\ & = \tan^{-1} \left(\dfrac{56}{33}\right) \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Prove that: $\;$ $\tan^{-1} \left(\dfrac{1}{5}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) - \tan^{-1} \left(\dfrac{1}{7}\right) = \tan^{-1} \left(\dfrac{21}{53}\right)$

$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1$

$\tan^{-1} \left(x\right) - \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right)$

$\begin{aligned} \tan^{-1} \left(\dfrac{1}{5}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) & = \tan^{-1} \left(\dfrac{\dfrac{1}{5} + \dfrac{1}{3}}{1 - \dfrac{1}{5} \times \dfrac{1}{3}}\right) \\\\ & = \tan^{-1} \left(\dfrac{8}{14}\right) \\\\ & = \tan^{-1} \left(\dfrac{4}{7}\right) \end{aligned}$

$\begin{aligned} \therefore \; \tan^{-1} \left(\dfrac{1}{5}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) - \tan^{-1} \left(\dfrac{1}{7}\right) & = \tan^{-1} \left(\dfrac{4}{7}\right) - \tan^{-1} \left(\dfrac{1}{7}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{4}{7} - \dfrac{1}{7}}{1 + \dfrac{4}{7} \times \dfrac{1}{7}}\right) \\\\ & = \tan^{-1} \left(\dfrac{21}{53}\right) \end{aligned}$

Hence proved.