JEE Main 2023 (Online) 13th April Evening ShiftQuestion 1
For $x\in(-1,1]$, the number of solutions of the equation $\sin^{-1}x=2\tan^{-1}x$ is equal to $\_\_\_\_\_\_$ .
Inverse Trigonometric Functions
Answer$2$
Solution
For $x\in(-1,1]$, $2\tan^{-1}x\in(-\frac\pi2,\frac\pi2]$, so taking $\sin$ is reversible: $x=\sin(2\tan^{-1}x)=\frac{2x}{1+x^2}$. $x=0$ or $1+x^2=2\Rightarrow x=\pm1$; $x=-1$ is excluded. Solutions: $x=0,1$. Answer $2$.