JEE Math Practice 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 :
- A.$${\pi \over 2}$$
- B.$${\pi \over 3}$$
- C.$${\pi \over 6}$$
- D.$${\pi \over 4}$$
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