Question

The value of $x$ for which $\sin \left({\cot }^{-1}(1+x)\right)=\cos \left({\tan }^{-1}x\right)$ is:

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. $-\frac{1}{2}$

Answer 1

Answer: (D)

$-\frac{1}{2}$

We have, $\sin \left({\cot }^{-1}(1+x)\right)=\cos \left({\tan }^{-1}x\right)$

$\Rightarrow \cos (\frac{\pi }{2}-{\cot }^{-1}(x+1))=\cos \left({\tan }^{-1}x\right)$

$\Rightarrow \frac{\pi }{2}-{\cot }^{-1}(x+1)=2n\pi \pm {\tan }^{-1}x$

Substitute $n=0\Rightarrow \frac{\pi }{2}-{\cot }^{-1}(x+1)=\pm {\tan }^{-1}x={\tan }^{-1}(\pm x)$

$\Rightarrow \frac{\pi }{2}={\tan }^{-1}(\pm x)+{\cot }^{-1}(x+1)$

$\Rightarrow x+1=\pm x\Rightarrow 2x+1=0\Rightarrow x=-\frac{1}{2}$