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$\tan ^{-1} \sqrt{3}-\cot ^{1}(-\sqrt{3})$ is equal to(A) $\pi$(B) -$\pi/2$(C) 0(D) $2 \sqrt{3}$
$\tan ^{-1} \sqrt{3}-\cot ^{1}(-\sqrt{3})$ is equal to
(A) $\pi$
(B) -$\pi/2$
(C) 0
(D) $2 \sqrt{3}$
CBSE | Class 12 | Inverse Trigonometric Functions | Maths | NCERT
$\tan ^{-1} \sqrt{3}-\cot ^{1}(-\sqrt{3})$ is equal to
(A) $\pi$
(B) -$\pi/2$
(C) 0
(D) $2 \sqrt{3}$

the correct Option is OPTION (B).

Reason:

$\tan ^{-1} \sqrt{3}-\cot ^{1}(-\sqrt{3})$ can be written as

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

$=\frac{\pi}{3}-\cot ^{-1}\left[\cot \left(\pi-\frac{\pi}{6}\right)\right]$

$=\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)$

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

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

$=-\pi / 2$

i

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