Question

please solve this question of MATHS....​

Answer 1

Answer:

$tan {}^{ - 1} \sqrt{3} - sec {}^{ - 1} ( - 2) = \\ \\ you \: know \: = = \\ = > sec {}^{ - 1} ( - x) = \pi - sec ^{ - 1} x \\ \\ \\ \\ \\ = > 60 - (\pi - 60) = 60 - \pi + 60 \\ \\ \\ = > \frac{\pi}{3} - (\pi - \frac{\pi}{3} ) \\ \\ \\ = > \frac{\pi}{3} - \pi + \frac{\pi}{3} \\ \\ \\ = > \frac{2\pi}{3} - \pi \\ \\ \\ = > \frac{2\pi - 3\pi}{3} = \\ \\ = > \frac{ - \pi}{3}$

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