JEE Main 2021 (Online) 20th July Evening ShiftQuestion 1
The value of $\tan\left(2\tan^{-1}\left(\dfrac{3}{5}\right)+\sin^{-1}\left(\dfrac{5}{13}\right)\right)$ is equal to :
Inverse Trigonometric Functions
Solution
$2\tan^{-1}\frac35=\tan^{-1}\frac{15}{8}$ and $\sin^{-1}\frac5{13}=\tan^{-1}\frac5{12}$. Value $=\dfrac{\frac{15}{8}+\frac5{12}}{1-\frac{15}{8}\cdot\frac5{12}}=\dfrac{55/24}{21/96}=\dfrac{220}{21}$. Option (B).