JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
If the inverse trignometric functions take principal values, then $\cos^{-1}\left(\frac{3}{10}\cos\left(\tan^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}\sin\left(\tan^{-1}\left(\frac{4}{3}\right)\right)\right)$ is equal to :
Inverse Trigonometric Functions
Solution
$\cos(\tan^{-1}\frac43)=\frac35$, $\sin(\tan^{-1}\frac43)=\frac45$. Argument $=\frac3{10}\cdot\frac35+\frac25\cdot\frac45=\frac{9}{50}+\frac{16}{50}=\frac12$. $\cos^{-1}\frac12=\frac\pi3$. Option (C).