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

JEE MainJEE Main 2023Inverse Trigonometric Functions

Solution:

$ \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)$
$ =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{3}$