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) $
$= \tan \left(\frac{2\theta}{2} + \frac{2\theta}{2}\right) =\tan \left(2\theta\right)=\tan \left(2 \tan^{-1} x\right)$
$= \tan \left(\tan^{-1} \left(\frac{2x}{1-x^{2}}\right) \right) =\frac{2x}{1-x^{2}}$