The principal value of $[\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})]$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $-\frac{\pi}{2}$ ##
We have,
$\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})$
$= \tan^{-1} \left( \tan \frac{\pi}{3} \right) - \left( \pi - \cot^{-1} \left( \cot \frac{\pi}{6} \right) \right)$
$= \frac{\pi}{3} - \left( \pi - \frac{\pi}{6} \right) = -\frac{\pi}{2}$ [Since $\cot^{-1}(-\theta) = \pi - \cot^{-1} \theta$]