if

$\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c}$

$\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\sin ^{-1} x+\cos ^{-1} x}{a+b}=\frac{\pi}{2(a+b)}$

Now, $\frac{\tan ^{-1} y}{c}=\frac{\pi}{2(a+b)}$

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

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