Question

Show that 2 tan⁻¹ x + sin⁻¹ $\frac{2x}{1+x^2}$ is constant for x ≥ 1, find that constant.

Answer 1

We have
$$\begin{gathered} 2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ \left(1\right)\mathrm{For}x>1, \\ =2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ =\mathrm{\pi }-{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\left[\because 2{\tan }^{-1}x=\mathrm{\pi }-{\sin }^{-1}\left(\frac{2x}{1+x^2}\right),x>1\right] \\ =\mathrm{\pi } \\ \left(2\right)\mathrm{For}x=1, \\ =2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ =2{\tan }^{-1}\left(1\right)+{\sin }^{-1}\left(\frac{2\left(1\right)}{1+{\left(1\right)}^2}\right) \\ =2{\tan }^{-1}\left(1\right)+{\sin }^{-1}\left(1\right) \\ =2\left(\frac{\mathrm{\pi }}{4}\right)+\frac{\mathrm{\pi }}{2} \\ =\mathrm{\pi } \end{gathered}$$