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If $\sin \left(\cot ^{-1}(x+1)=\cos \left(\tan ^{-1} x\right)\right.$, then $x=$
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Verified Answer
The correct answer is:
$-\frac{1}{2}$
$\sin \left[\cot ^{-1}(x+1)\right]=\sin \left(\sin ^{-1} \frac{1}{\sqrt{x^2+2 x+2}}\right)$
$=\frac{1}{\sqrt{x^2+2 x+2}}$
$\begin{aligned} & \cos \left(\tan ^{-1} x\right)=\cos \left(\cos ^{-1} \frac{1}{\sqrt{1+x^2}}\right)=\frac{1}{\sqrt{1+x^2}} \\ & \text { Thus, } \frac{1}{\sqrt{x^2+2 x+2}}=\frac{1}{\sqrt{1+x^2}} \\ & \Rightarrow x^2+2 x+2=1+x^2 \Rightarrow x=-\frac{1}{2} .\end{aligned}$
$=\frac{1}{\sqrt{x^2+2 x+2}}$
$\begin{aligned} & \cos \left(\tan ^{-1} x\right)=\cos \left(\cos ^{-1} \frac{1}{\sqrt{1+x^2}}\right)=\frac{1}{\sqrt{1+x^2}} \\ & \text { Thus, } \frac{1}{\sqrt{x^2+2 x+2}}=\frac{1}{\sqrt{1+x^2}} \\ & \Rightarrow x^2+2 x+2=1+x^2 \Rightarrow x=-\frac{1}{2} .\end{aligned}$
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