Question

If $x_1=2ta{n}^{-1}\left(\frac{1+x}{1-x}\right),x_2=si{n}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, where $x_1,x_2ϵ(0,1)$, then $2(x_1+x_2)$ is equal to

• A. $0$
• B. $2\pi$
• C. $\pi$
• D. $noneofthese$

Answer 1

The correct option is C $\pi$

$x_1=2{\tan }^{-1}\left(\frac{1+x}{1-x}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)x_2={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Therefore,
$x_1+x_2={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\pi }{2}\Rightarrow 2(x_1+x_2)=\pi$