Question

$\tan \left[\frac{1}{2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+\frac{1}{2}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]=$

• A. $\infty$
• B. 1
• C. $\frac{2x}{1-x^2}$
• D. $\frac{2x}{1+x^2}$

Answer 1

Answer: (C)

$\tan \left[\frac{1}{2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+\frac{1}{2}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$
Putting $x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$
$=\tan \left[\frac{1}{2}{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)+\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)\right]$
$=\tan \left(\frac{1}{2}{\sin }^{-1}\left(\sin 2\theta \right)+\frac{1}{2}{\cos }^{-1}\left(\cos 2\theta \right)\right)$
$=\tan \left(\frac{2\theta }{2}+\frac{2\theta }{2}\right)=\tan \left(2\theta \right)=\tan \left(2{\tan }^{-1}x\right)$
$=\tan \left({\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\right)=\frac{2x}{1-x^2}$