# 21. $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$  is equal to        (A)    $\pi$        (B)    $-\frac{\pi}{2}$        (C)    0        (D)    $2\sqrt3$

We have $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$;

finding the value of  $\cot^{-1}(-\sqrt3)$:

Assume $\cot^{-1}(-\sqrt3) =y$ then,

$\cot y = -\sqrt 3$  and the range of the principal value of $\cot^{-1}$ is $(0,\pi)$.

Hence, principal value is $\frac{5\pi}{6}$

Therefore $\cot^{-1} (-\sqrt3) = \frac {5\pi}{6}$

and $\tan^{-1} \sqrt3 = \frac{\pi}{3}$

so, we have now,

$\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)=\frac{\pi}{3} - \frac{5\pi}{6}$

$= \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6}$

or, $= \frac{ -\pi}{2}$

Hence the answer is option  (B).

Solve the principal values for them seperately, and you will get option; (B)

$= \frac{ -\pi}{2}$

Hence, (B) is the right answer

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