Question

If $0\le x\le 1$, then $\tan \left\{\frac{1}{2}{\sin }^{-1}\frac{2x}{1+x^2}+\frac{1}{2}{\cos }^{-1}\frac{2x}{1+x^2}\right\}$

• A. 1
• B. $0$
• C. $\frac{2x}{1+x^2}$
• D. $x$

Answer 1

Answer: (A)

1

$tan\left[\frac{1}{2}\right(si{n}^{-1}\left(\frac{2x}{1+x^2}\right)+co{s}^{-1}\left(\frac{2x}{1+x^2}\right)\left)\right]$
$=tan\left[\frac{1}{2}\right(si{n}^{-1}\left(\theta \right)+co{s}^{-1}\left(\theta \right)\left)\right]$
$=tan\left(\frac{\pi }{2\left(2\right)}\right)$
$=tan\left(\frac{\pi }{4}\right)$
$=1$