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Question

${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$ is equal to :

${\pi \over 2}$
${\pi \over 3}$
${\pi \over 6}$
${\pi \over 4}$

Solution

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