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

The correct option is A $1$

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 $0\le x<\infty$
Therefore, $\sin \left\{{\tan }^{-1}\frac{1-x^2}{2x}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \left\{{\cot }^{-1}\frac{2x}{1-x^2}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \left\{\frac{\pi }{2}-{\tan }^{-1}\frac{2x}{1-x^2}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\right\}$
$=\sin \left\{\frac{\pi }{2}-2{\tan }^{-1}x+2{\tan }^{-1}\right\},$ if $0\le x<1$
$=\sin \frac{\pi }{2}=1$