JEE Main 2023 (Online) 24th January Morning ShiftQuestion 1
$\tan^{-1}\frac{1+\sqrt3}{3+\sqrt3}+\sec^{-1}\sqrt{\frac{8+4\sqrt3}{6+3\sqrt3}}=$
Inverse Trigonometric Functions
Solution
$\frac{1+\sqrt3}{3+\sqrt3}=\frac1{\sqrt3}$ gives $\frac\pi6$; $\sqrt{\frac{4(2+\sqrt3)}{3(2+\sqrt3)}}=\frac2{\sqrt3}$ gives $\sec^{-1}\frac2{\sqrt3}=\frac\pi6$. Sum $=\frac\pi3$.