A value of $x$ for which $\sin \left(\cot ^{-1}(1+x) ight)=\cos$ $\left( an ^{-1} x ight)$, is :

Question

A value of $x$ for which $\sin \left(\cot ^{-1}(1+x)ight)=\cos$ $\left(	an ^{-1} xight)$, is :

Aicharya · Accepted Answer

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

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

Solution