Kerala KEAM 5 June 2024 Solved Paper

We are given the expression $\sin \left( 2 \sin^{-1} x + \cos^{-1} x \right)$, and we need to simplify it.Let $\theta = \sin^{-1} x$. Then $\sin \theta = x$, and the angle $\theta$ lies within the range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$. Also, recall that: $$\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x = \frac{\pi}{2} - \theta$$Now, we can rewrite the expression as: $$\sin \left( 2 \theta + \left( \frac{\pi}{2} - \theta \right) \right)$$This simplifies to: $$\sin \left( \theta + \frac{\pi}{2} \right)$$Using the trigonometric identity $\sin \left( \theta + \frac{\pi}{2} \right) = \cos \theta$, we have: $$\sin \left( 2 \sin^{-1} x + \cos^{-1} x \right) = \cos \theta$$Since $\sin \theta = x$, we use the Pythagorean identity $\cos^2 \theta = 1 - \sin^2 \theta$ to find: $$\cos \theta = \sqrt{1 - x^2}$$Thus, the value of $\sin \left( 2 \sin^{-1} x + \cos^{-1} x \right)$ is $\sqrt{1 - x^2}$.Thus, the correct answer is option (A), $\sqrt{1 - x^2}$. Quick Tip: Use the identities $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $\cos(2A) = 1 - 2 \sin^2 A$ to simplify trigonometric expressions involving inverse functions.