If $\frac{{\sin }^{-1}x}{a}=\frac{{\cos }^{-1}x}{b}=\frac{{\tan }^{-1}y}{c}$ then the value of  $\cos \left(\frac{\pi c}{a+b}\right)$ is

Let ${\sin }^{-1}x=\theta \text{.}$ Then $, {\cos }^{-1}x=\frac{\pi }{2}-\theta \text{. }$

$\begin{array}{l}\therefore \frac{{\sin }^{-1}x}{a}=\frac{{\cos }^{-1}x}{b}=\frac{{\tan }^{-1}y}{c}\\ \Rightarrow \frac{\theta }{a}=\frac{\frac{\pi }{2}-\theta }{b}=\frac{{\tan }^{-1}y}{c} \left[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\right]\end{array}$

$\Rightarrow \frac{\theta }{a}=\frac{\pi }{2b}-\frac{\theta }{b}$ and $\frac{\theta }{a}=\frac{{\tan }^{-1}y}{c}$

$\Rightarrow \theta \left(\frac{a+b}{ab}\right)=\frac{\pi }{2b}$ and $\theta =\frac{a}{c}{\tan }^{-1}y$

$\Rightarrow \frac{a}{c}\left(\frac{a+b}{ab}\right){\tan }^{-1}y=\frac{\pi }{2b} \text{ On eliminating }\theta ]$

$\begin{array}{l}\Rightarrow 2{\tan }^{-1}y=\frac{\pi c}{(a+b)}\\ \Rightarrow {\cos }^{-1}\left(\frac{1-y^2}{1+y^2}\right)=\frac{\pi c}{a+b}\left[\because 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]\\ \Rightarrow \frac{1-y^2}{1+y^2}=\cos \left(\frac{\pi c}{a+b}\right)\end{array}$