Question

If x > 1, then write the value of sin⁻¹$\left(\frac{2x}{1+x^2}\right)$ in terms of tan⁻¹ x.

Answer 1

$$\begin{gathered} {\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ =\mathrm{\pi }-2{\tan }^{-1}x\left[\because 2{\tan }^{-1}x=\mathrm{\pi }-{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\mathrm{for}\mathrm{x}>1\right] \end{gathered}$$