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

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Inverse Trigonometric Functions

tan^{- 1} (\frac{1 + 3}{3 + 3}) + sec^{- 1} (\frac{8 + 4 3}{6 + 3 3})

is equal to

Answer: (C)

Rationalising,

\frac{1 + 3}{3 + 3} = \frac{1 + 3}{3 ( 3 + 1 )} = \frac{1}{3}

, so the first term is

tan^{- 1} \frac{1}{3} = \frac{π}{6}

. Also

\frac{8 + 4 3}{6 + 3 3} = \frac{4 ( 2 + 3 )}{3 ( 2 + 3 )} = \frac{4}{3}

, so the second term is

sec^{- 1} \frac{2}{3} = cos^{- 1} \frac{3}{2} = \frac{π}{6}

. Their sum is

\frac{π}{3} .

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