Question

The value of $-4x$ for which $\sin \left[{\cot }^{-1}\right(1+x\left)\right]=\cos \left({\tan }^{-1}x\right)$ is _______.

Answer 1

$\begin{array}{c}\Rightarrow \sin \left[{\cot }^{-1}\right(1+x\left)\right]=\sin (\frac{\pi }{2}-{\tan }^{-1}x)\\ \Rightarrow \left[{\cot }^{-1}\right(1+x\left)\right]=(\frac{\pi }{2}-{\tan }^{-1}x)\\ \Rightarrow {\tan }^{-1}\frac{1}{\left(1+x\right)}=\frac{\pi }{2}{\tan }^{-1}x\\ \Rightarrow {\tan }^{-1}\left[\frac{\frac{1}{1+x}+x}{1-\frac{x}{\left(1+x\right)}}\right]=\frac{\pi }{2}\\ \Rightarrow \frac{\frac{x^2+x+1}{\left(1+x\right)}}{\frac{1}{\left(1+x\right)}}=\tan \frac{\pi }{2}\\ \Rightarrow x^2+x+1=\infty \\ itsrootsisx=\frac{-\left(1+\sqrt{3}\right)}{2}or\\ x=\frac{\left(\sqrt{3}-1\right)}{2}\\ therefore-4x=-4\times -\frac{\left(1+\sqrt{3}\right)}{2}\end{array}$