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.