Question
$\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})$ is equal to
  1. $\pi$
  2. $-\frac{\pi}{2}$
  3. 0
  4. $2\sqrt{3}$

Answer

$\tan^{-1}\sqrt{3}-\cot^{-1}\bigg(-\sqrt{3}\bigg)=\tan^{-1}\tan\frac{\pi}{3}-\cot^{-1}\bigg(-\cot\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\cot^{-1}\bigg[\cot\bigg(\pi-\frac{\pi}{6}\bigg)\bigg]=\frac{\pi}{3}-\bigg(\pi-\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}$
Therefore, option (b) is correct.

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