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

• A

  \(\frac{1}{2}\)
• B

  1
• C

  0
• Correct answer
  D

  \(-\frac{1}{2}\)

Solution

\(\begin{aligned} & \sin \left(\cot ^{-1}(1+x)\right)=\frac{1}{\sqrt{2+2 x+x^2}} \\ & \cos \left(\tan ^{-1} x\right)=\frac{1}{\sqrt{1+x^2}} \end{aligned}\) So, \(\frac{1}{\sqrt{2+2 x+x^2}}=\frac{1}{\sqrt{1+x^2}}\) \(\Rightarrow 1+2 \mathrm{x}=0 \Rightarrow \mathrm{x}=-\frac{1}{2}\)

Hence, the correct answer is option: 'D'.