Question

cot inverse ( 1 - x / 1 + x )

Answer 1

$$\begin{gathered} {\cot }^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right) \\ \mathrm{let}\mathrm{x}=\tan \mathrm{y} \\ {\cot }^{-1}\left(\frac{1-\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)\tan \mathrm{y}}{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan \mathrm{y}}\right)={\cot }^{-1}\left(\frac{1}{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}+\mathrm{y}\right)}\right)\mathrm{since}\tan \left(\mathrm{x}+\mathrm{y}\right)=\frac{\tan \mathrm{x}+\mathrm{tany}}{1-\tan \mathrm{x}\tan \mathrm{y}} \\ ={\cot }^{-1}\left(\cot \left(\frac{\mathrm{\pi }}{4}+\mathrm{y}\right)\right)=\frac{\mathrm{\pi }}{4}+\mathrm{y} \\ =\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\mathrm{x} \\ \mathrm{On}\mathrm{differentiating}\mathrm{wrt}\mathrm{x},\mathrm{we}\mathrm{get} \\ \frac{\mathrm{d}\left(\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\mathrm{x}\right)}{\mathrm{dx}}=\frac{1}{1+{\mathrm{x}}^2}\mathrm{ANS}... \end{gathered}$$