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14. $\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$   is equal to

(A)       $\pi$

(B)   $-\frac{\pi}{3}$

(C)       $\frac{\pi}{3}$

(D)     $\frac{2\pi}{3}$

Answers (1)

best_answer

Let us assume the values of $\tan^{-1}(\sqrt3)$ be 'x' and $\sec^{-1}(-2)$ be 'y'.

Then we have;

$\tan^{-1}(\sqrt3) = x$ or $\tan x = \sqrt 3$ or $\tan \frac{\pi}{3} = \sqrt 3$ or

$x = \frac{\pi}{3}$ .

and $\sec^{-1}(-2) = y$ or $\sec y = -2$

or $-\sec (\frac{\pi}{3}) =\sec ({\pi - \frac{\pi}{3}}) = \sec{\frac{2\pi}{3}}$

$y = \frac{2\pi}{3}$

also the ranges of the principal values of $\tan^{-1}$ and $\sec^{-1}$ are $(\frac{-\pi}{2},\frac{\pi}{2})$ . and

$[0,\pi] - \left \{ \frac{\pi}{2} \right \}$ respectively.

$\therefore$ we have then;

$\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$

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

Posted by

Divya Prakash Singh

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