JEE Main 2021 (Online) 26th February Morning ShiftQuestion 1
If $\frac{\sin^{-1}x}{a}=\frac{\cos^{-1}x}{b}=\frac{\tan^{-1}y}{c}$; $0<x<1$, then the value of $\cos\left(\frac{\pi c}{a+b}\right)$ is :
Inverse Trigonometric Functions
Solution
Since $\sin^{-1}x+\cos^{-1}x=\frac\pi2$, with common ratio $k$: $k(a+b)=\frac\pi2$. $\tan^{-1}y=ck=\frac{\pi c}{2(a+b)}\Rightarrow\frac{\pi c}{a+b}=2\tan^{-1}y$. $\cos(2\tan^{-1}y)=\frac{1-y^2}{1+y^2}$. Option (C).