Question

Given that ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\{\begin{array}{ll}2{\tan }^{-1}x,& ,|x|<1\\ -\pi +2{\tan }^{-1}x,& x>1,\\ \pi +2{\tan }^{-1}x& ,x<-1\end{array}$ ,${\sin }^{-1}\frac{2x}{1+x^2}=\{\begin{array}{llc}2{\tan }^{-1}x& \text{ if }& |x|\le 1\\ -\pi -2{\tan }^{-1}x& \text{ if }& x>1\\ -\left(\pi +2{\tan }^{-1}x\right)& \text{ if }& x<-1\end{array}$ and ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ for $-1\le x\le 1$
If $(x-1)\left(x^2+1\right)>0$, then $\sin \left(\frac{1}{2}{\tan }^{-1}\frac{2x}{1-x^2}-{\tan }^{-1}x\right)=$

• A. 1
• B. $\frac{1}{\sqrt{2}}$
• C. -1
• D. None of these

Answer 1

Answer: (C)

Correct answer is (c) -1