Question

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

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $0$
• D. $\frac{9}{4}$

Answer 1

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

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

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

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

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

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

$\Rightarrow x^2+1=x^2+2x+2$

$\Rightarrow 2x+1=0$

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