Question

Write the principal value of tan^-1(√3) - cot^-1(-√3).

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{range}\mathrm{of}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\tan }^{-1}\mathrm{is}\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right). \\ \mathrm{Let}{\tan }^{-1}\left(\sqrt{3}\right)=\mathrm{\alpha } \\ \Rightarrow \tan \mathrm{\alpha }=\sqrt{3} \\ \Rightarrow \tan \mathrm{\alpha }=\tan \left(\mathrm{\pi }/3\right) \\ \Rightarrow \mathrm{\alpha }=\frac{\mathrm{\pi }}{3}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right). \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{range}\mathrm{of}\mathrm{principal}\mathrm{value}\mathrm{branch}{\mathrm{ofcot}}^{-1}\mathrm{is}\left(0,\mathrm{\pi }\right). \\ \mathrm{Let}{\cot }^{-1}\left(-\sqrt{3}\right)=\mathrm{\beta } \\ \Rightarrow \cot \mathrm{\beta }=-\sqrt{3} \\ \Rightarrow \cot \mathrm{\beta }=-\cot \left(\mathrm{\pi }/6\right) \\ \Rightarrow \cot \mathrm{\beta }=\cot \left[\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right] \\ \Rightarrow \cot \mathrm{\beta }=\cot \left[\frac{5\mathrm{\pi }}{6}\right] \\ \Rightarrow \mathrm{\alpha }=\frac{5\mathrm{\pi }}{6}\in \left(0,\mathrm{\pi }\right). \\ \mathrm{Now},\mathrm{principal}\mathrm{value}\mathrm{of}{\tan }^{-1}\left(\sqrt{3}\right)-{\cot }^{-1}\left(-\sqrt{3}\right)=\frac{\mathrm{\pi }}{3}-\frac{5\mathrm{\pi }}{6}=-\frac{3\mathrm{\pi }}{6}=-\frac{\mathrm{\pi }}{2} \end{gathered}$$