Q.
Given that $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\begin{cases}2 \tan ^{-1} x, & ,|x|<1 \\ -\pi+2 \tan ^{-1} x, & x>1, \\ \pi+2 \tan ^{-1} x & , x<-1\end{cases}$ ,$\sin ^{-1} \frac{2 x}{1+x^2}=\begin{cases}2 \tan ^{-1} x & \text { if } & |x| \leq 1 \\ -\pi-2 \tan ^{-1} x & \text { if } & x>1 \\ -\left(\pi+2 \tan ^{-1} x\right) & \text { if } & x<-1\end{cases}$
and $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ for $-1 \leq x \leq 1$
If $(x-1)\left(x^2+1\right)>0$, then $\sin \left(\frac{1}{2} \tan ^{-1} \frac{2 x}{1-x^2}-\tan ^{-1} x\right)=$
Inverse Trigonometric Functions
Solution: