Question

$\sin\left[\tan^{-1}\frac{1-x^2}{2x}+\cos^{-1}\frac{1-x^2}{1+x^2}\right]$ is equal to

• A. 0
• B. 1
• C. $\frac{1}{\sqrt2}$
• D. $\sqrt2$

Answer 1

Answer: (B)

Put $x= tan \theta$
$ \therefore sin \left[tan^{-1} \frac{1-x^{2}}{2x} + cos^{-1} \frac{1-x^{2}}{1+x^{2}}\right] $
$=sin\left[tan^{-1}\left(cot\,2\,\theta\right)+ cos^{-1} cos\, 2\,\theta\right]$
$ = sin \left[tan^{-1} tan\left(\frac{\pi}{2} -2\,\theta\right)+2\,\theta\right] $
$ = sin\left(\frac{\pi}{2}-2\,\theta+2\,\theta\right)$
$ = sin \frac{\pi}{2} = 1$