Question

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

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

Answer 1

The correct option is A $-\frac{1}{2}$

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

apply $\cos (\frac{\pi }{2}-\theta )=\sin \theta$

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

apply ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$

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

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

apply ${\tan }^{-1}(-x)=-{\tan }^{-1}\left(x\right)$

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

$\Rightarrow {\tan }^{-1}(-x)={\tan }^{-1}(1+x)$

$\Rightarrow -x=1+x$

$\Rightarrow x=-\frac{1}{2}$