^ -1 ( 1+ 3 3+ 3 )+ ^ -1 ( 8+4 3 6+3 3 ) is equal to
Mathematics · JEE Main · NTA Exams — Trigonometry
\[\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)\] is equal to
Answer
(A)
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