If in any $\triangle ABC$, $a = 5$, $b = 4$ and $\cos \left(A - B\right) = \dfrac{31}{32}$, then show that the third side of the triangle is $c = 6$.
In any $\triangle ABC$, by law of tangents,
$\tan \left(\dfrac{A - B}{2}\right) = \dfrac{a - b}{a + b} \cot \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (1)$
Now, by definition,
$\cot \left(\dfrac{C}{2}\right) = \dfrac{\sin C}{1 - \cos C}$ $\;\;\; \cdots \; (2a)$,
$\tan \left(\dfrac{A - B}{2}\right) = \dfrac{1 - \cos \left(A - B\right)}{\sin \left(A - B\right)}$ $\;\;\; \cdots \; (2b)$
and $\;$ $\sin \left(A - B\right) = \sqrt{1 - \cos^2 \left(A - B\right)}$ $\;\;\; \cdots \; (2c)$
$\therefore \;$ In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes
$\dfrac{1 - \cos \left(A - B\right)}{\sqrt{1 - \cos^2 \left(A - B\right)}} = \left(\dfrac{a - b}{a + b}\right) \left(\dfrac{\sin C}{1 - \cos C}\right)$ $\;\;; \cdots \; (3)$
Given: $\;$ $a = 5$, $\;$ $b = 4$, $\;$ $\cos \left(A - B\right) = \dfrac{31}{32}$ $\;\;\; \cdots \; (4)$
$\therefore \;$ In view of equation $(4)$ equation $(3)$ becomes,
$\dfrac{1 - \dfrac{31}{32}}{\sqrt{1 - \dfrac{961}{1024}}} = \left(\dfrac{5 - 4}{5 + 4}\right) \left(\dfrac{\sin C}{1 - \cos C}\right)$
i.e. $\;$ $\dfrac{\dfrac{1}{32}}{\dfrac{\sqrt{63}}{32}} = \dfrac{1}{9} \times \left(\dfrac{\sqrt{1 - \cos^2 C}}{1 - \cos C}\right)$
i.e. $\;$ $\dfrac{\sqrt{1 - \cos^2 C}}{1 - \cos C} = \dfrac{9}{\sqrt{63}}$
i.e. $\;$ $\dfrac{1 - \cos^2 C}{\left(1 - \cos C\right)^2} = \dfrac{81}{63}$
i.e. $\;$ $\dfrac{\left(1 + \cos C\right) \left(1 - \cos C\right)}{\left(1 - \cos C\right)^2} = \dfrac{81}{63}$
i.e. $\;$ $\dfrac{1 + \cos C}{1 - \cos C} = \dfrac{81}{63}$ $\;\;\;$ provided $\;\;$ $\left(1 - \cos C\right) \neq 0$
i.e. $\;$ $63 + 63 \cos C = 81 - 81 \cos C$
i.e. $\;$ $144 \cos C = 18$
$\implies$ $\cos C = \dfrac{1}{8}$ $\;\;\; \cdots \; (5)$
By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$ $\;\;\; \cdots \; (6)$
In view of equations $(4)$ and $(5)$ equation $(6)$ becomes,
$\dfrac{1}{8} = \dfrac{25 + 16 - c^2}{2 \times 5 \times 4}$
i.e. $\;$ $5 = 41 - c^2$
i.e. $\;$ $c^2 = 36$ $\implies$ $c = 6$