JEE MAIN - Mathematics Hindi (2023 - 24th January Morning Shift - No. 5)

$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$ बराबर है
$$\frac{\pi}{2}$$
$$\frac{\pi}{3}$$
$$\frac{\pi}{6}$$
$$\frac{\pi}{4}$$

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