Question

The value of $\tan \{\frac{1}{2}{\sin }^{-1}\frac{2x}{1+x^2}+\frac{1}{2}{\cos }^{-1}\frac{1-x^2}{1+x^2}\}$ is

• A. $\frac{2x}{1+x^2}$ if $0\le x<1$
• B. $\frac{2x}{1+x^2}$ if $x<1$
• C. not finite if $x>1$
• D. none of these

Answer 1

Answer: (A), (C)

$\frac{2x}{1+x^2}$ if $0\le x<1$
not finite if $x>1$

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