If $\sin \left(\cot ^{-1}(x+1) ight)=\cos \left( an ^{-1} x ight)$ then considering positive square roots, $x$ has the value $\qquad$

Question

If $\sin \left(\cot ^{-1}(x+1)ight)=\cos \left(	an ^{-1} xight)$ then considering positive square roots, $x$ has the value $\qquad$

Aicharya · Accepted Answer

$\begin{aligned} & \sin \left(\cot ^{-1}(x+1)ight)=\cos \left(	an ^{-1} xight) \ & \sin \left[\sin ^{-1} \frac{1}{\sqrt{x^2+2 x+2}}ight]=\cos \left[\cos ^{-1} \frac{1}{\sqrt{1+x^2}}ight] \ & \frac{1}{\sqrt{x^2+2 x+2}}=\frac{1}{\sqrt{1+x^2}} \ & \Rightarrow \sqrt{1+x^2}=\sqrt{x^2+2 x+2} \ & \Rightarrow 1+x^2=x^2+2 x+2 \ & \Rightarrow 2 x+1=0 \ & \Rightarrow x=\frac{-1}{2}\end{aligned}$

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

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