Question

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

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

Answer 1

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

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

${\cot }^{1-}(1+x)={\sin }^{-1}\frac{1}{\sqrt{1+(x+1{)}^2}}$

$[\because {\tan }^{-1}x={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}]$

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

$\frac{1}{\sqrt{x^2+2x+2}}=\frac{1}{1+x^2}$

$x^2+1=x^2+2x+2$

$2x+1=0$

$x=-1/2$.