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)

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}

 

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