Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. None of the above

Answer 1

Answer: (A)

We know that, ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=2{\tan }^{-1}x,$
if $-1<x<1$ and ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x,$
if $0\le x\le \infty$
$\therefore$ $\sin \left\{{\tan }^{-1}\frac{1-x^2}{2x}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \left\{{\cot }^{-1}\frac{2x}{1-x^2}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \left\{\frac{\pi }{2}-{\tan }^{-1}\frac{2x}{1-x^2}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \left\{\frac{\pi }{2}-2{\tan }^{-1}x+2{\tan }^{-1}x\right\}$ ,
if $0\le x<1$
$=\sin \frac{\pi }{2}=1$