JEE PYQ: Inverse Trigonometric Functions Question 16
Question 16 - 2019 (12 Apr Shift 1)
The value of $\sin^{-1}\left(\frac{12}{13}\right) - \sin^{-1}\left(\frac{3}{5}\right)$ is equal to:
(1) $\pi - \sin^{-1}\left(\frac{63}{65}\right)$
(2) $\frac{\pi}{2} - \sin^{-1}\left(\frac{56}{65}\right)$
(3) $\frac{\pi}{2} - \cos^{-1}\left(\frac{9}{65}\right)$
(4) $\pi - \cos^{-1}\left(\frac{33}{65}\right)$
Show Answer
Answer: (2)
Solution
$\sin^{-1}\frac{12}{13} - \sin^{-1}\frac{3}{5} = -\sin^{-1}\left(\frac{12}{13}\cdot\frac{4}{5} - \frac{3}{5}\cdot\frac{5}{13}\right)$… Using the identity: $\sin^{-1}\frac{12}{13} - \sin^{-1}\frac{3}{5} = \sin^{-1}\left(\frac{12}{13}\cdot\frac{4}{5} - \frac{5}{13}\cdot\frac{3}{5}\right)$. Since $xy \ge 0$ and $x^2 + y^2 \le 1$ doesn’t hold, we use: $= \sin^{-1}\left(\frac{33}{65}\right) = \cos^{-1}\left(\frac{56}{65}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{56}{65}\right)$.
BETA
