Question
Write the principal value of $\tan^{-1}(\sqrt{3}) - \cot^{-1}( - \sqrt{3}).$

Answer

$\tan^{-1}\sqrt{3} - \cot^{-1}(- \sqrt{3}) = \tan^{-1}\bigg(\tan\frac{\pi}{3}\bigg) -\cot^{-1}\bigg(- \cot\frac{\pi}{6}\bigg)$
$ =\tan^{-1}\bigg(\tan\frac{\pi}{3}\bigg) - \cot^{-1}\bigg(\cot\bigg(\pi - \frac{\pi}{6}\bigg)\bigg)$
$ = \tan^{-1}\bigg(\tan\frac{\pi}{3}\bigg) - \cot^{-1}\bigg(\cot\frac{5\pi}{6}\bigg)$
$ = \frac{\pi}{3} - \frac{5\pi}{6}\bigg[\because\frac{\pi}{3}\in\bigg( -\frac{\pi}{2} , \frac{\pi}{2}\bigg)\text{ and }\frac{5\pi}{6}\in(0,\pi)\bigg]$
$ = \frac{2\pi - 5 \pi}{6} = - \frac{\pi}{2}.$

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