Question

If x > 1, then $2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$is equal to
(a) $4{\tan }^{-1}x$
(b) 0
(c) $\frac{\mathrm{\pi }}{2}$
(d) $\mathrm{\pi }$

Answer 1

$$\begin{gathered} 2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x+2{\tan }^{-1}x\left[\because {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x\right] \\ =4{\tan }^{-1}x \end{gathered}$$
Hence, the correct answer is option (a)