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

Question

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

Aicharya · Accepted Answer

Note that $\cot ^{-1} x=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}ight)$ and $\begin{aligned} & 	an ^{-1}(1+x)=\cos ^{-1}\left(\frac{1}{\sqrt{1+(1+x)^2}}ight) \ 	herefore \quad & \sin \left(\cot ^{-1}(x)ight)=\cos \left(	an ^{-1}(1+x)ight) \ \Rightarrow & \sin \left(\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}ight)ight)=\cos \left(\cos ^{-1}\left(\frac{1}{\sqrt{1+(1+x)^2}}ight)ight) \ \Rightarrow & \frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+(1+x)^2}} \ \Rightarrow & 1+(1+x)^2=1+x^2 \ \Rightarrow & x=\frac{-1}{2} \end{aligned}$

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

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