Question

tan^-1(√3)+sec^-1(-2)

Answer 1

Answer:

Your question can be simplified as  $\frac{\sqrt{3}}{tan} +\frac{-2}{sec} = \frac{\pi}{3}+\frac{\pi}{3} = \frac{2\pi}{3}$

Step-by-step explanation:

So let $\frac{\sqrt{3}}{tan} =x,\frac{2}{sec} =y\\\\Hence\\  tan (x) = \sqrt{3} = tan (\frac{\pi}{3})\\ sec (y)= 2 = \frac{1}{cos(y)}=\frac{1}{cos(\frac{\pi}{3}) }  =sec(\frac{\pi}{3} )$

Thus

$\frac{\sqrt{3}}{tan} +\frac{-2}{sec} = \frac{\pi}{3}+\frac{\pi}{3} = \frac{2\pi}{3}$

Hope it helped

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