Question

Solve $\sin \left[{\cot }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]=$

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

Answer 1

Answer: (B)

$1$

Given $\sin \left[{\cot }^{-1}\right(\frac{2x}{1-x^2})+{\cos }^{-1}\frac{(1-x^2)}{(1+x^2)}]$

Let $x=\tan \theta$

$\Rightarrow \sin \left({\cot }^{-1}\right(\frac{2\tan \theta }{1-{\tan }^2\theta })+{\cos }^{-1}(\frac{(1-{\tan }^2\theta }{1+{\tan }^2\theta )}]$

$\Rightarrow \sin \left({\cot }^{-1}\right(\tan 2\theta )+{\cos }^{-1}(\cos 2\theta \left)\right]$

$\Rightarrow [\frac{\pi }{2}-{\tan }^{-1}(\tan 2\theta )+2\theta ]$

$\Rightarrow \sin [\frac{\pi }{2}-2\theta +2\theta ]$

$\Rightarrow 1$