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tan inverse root 3 - cot inverse (- root 3) is equal to (A) pi (B) - pi / 2 (C) 0 (D) 2 root 3

21. \tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)  is equal to

        (A)    \pi

        (B)    -\frac{\pi}{2}

        (C)    0

        (D)    2\sqrt3

Answers (1)
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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).

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