Question

If $-1\le x\le 0$, then $\sin \{{\tan }^{-1}\frac{1-x^2}{2x}-{\cos }^{-1}\frac{1-x^2}{1+x^2}\}$ is equal to

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The correct option is B -1

${\tan }^{-1}x=-\pi +{\cot }^{-1}\left(\frac{1}{x}\right)fix<0$
$\therefore \sin \left\{{\tan }^{-1}\frac{1-x^2}{2x}-{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \{-\pi +{\cot }^{-1}\frac{2x}{1-x^2}-{\cos }^{-1}\frac{1-x^2}{1+x^2}\}$
$=\sin \{-\pi +\frac{\pi }{2}-{\tan }^{-1}\frac{2x}{1-x^2}-{\cos }^{-1}\frac{1-x^2}{1+x^2}\}$
$=\sin \{-\frac{\pi }{2}-2{\tan }^{-1}x+2{\tan }^{-1}x\},[\because {\cos }^{-1}\frac{1-x^2}{1+x^2}=-2{\tan }^{-1}xifx<0]$
$=\sin (-\frac{\pi }{2})=-1$