Prove that tan[(inverse) x ] + cot[(inverse) x ] = pi/2.

Dear Student,

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.={\tan }^{-1}\mathrm{x}+{\cot }^{-1}\mathrm{x} \\ ={\tan }^{-1}\mathrm{x}++{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right) \\ ={\tan }^{-1}\left[\frac{\mathrm{x}+\left(\frac{1}{\mathrm{x}}\right)}{1-\mathrm{x}\left(\frac{1}{\mathrm{x}}\right)}\right] \\ ={\tan }^{-1}\left[\frac{\mathrm{x}+\left(\frac{1}{\mathrm{x}}\right)}{0}\right] \\ ={\tan }^{-1}\left(\infty \right) \\ ={\tan }^{-1}\left(\tan \frac{\mathrm{\pi }}{2}\right) \\ =\frac{\mathrm{\pi }}{2} \\ =\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \mathrm{Hence} \mathrm{Proved}. \end{gathered}$$

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