Question

Given \[ 0 \le x \le \frac{1}{2}, \] find the value of \[ \tan \left( \sin^{-1}\left( \frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^2}}{\sqrt{2}} \right) - \sin^{-1} x \right). \] 

• A. 1
• B. $\sqrt{3}$
• C. -1
• D. $\frac{1}{\sqrt{3$

Hint:

When dealing with expressions involving $\sin^{-1}(A \cos B \pm C \sin D)$, try to convert the argument of the inverse sine into a single sine term using trigonometric sum/difference identities. Pay attention to the domain of the inverse function to correctly apply $\sin^{-1}(\sin x) = x$.

Answer 1

The Correct Option is A

Step 1: Let the expression be: \[ E = \tan \left( \sin^{-1}\left( \frac{x}{\sqrt{2}} + \frac{\sqrt{1-x^2}}{\sqrt{2}} \right) - \sin^{-1} x \right) \]
Step 2: Substitute: \[ \sin^{-1} x = \theta \Rightarrow x = \sin\theta \] Given: \[ 0 \le x \le \frac{1}{2} \Rightarrow 0 \le \theta \le \frac{\pi}{6} \] \[ \sqrt{1-x^2} = \cos\theta \]
Step 3: Substitute into expression: \[ E = \tan \left( \sin^{-1}\left( \frac{\sin\theta}{\sqrt{2}} + \frac{\cos\theta}{\sqrt{2}} \right) - \theta \right) \]
Step 4: Use identity: \[ \frac{1}{\sqrt{2}} = \sin\frac{\pi}{4} = \cos\frac{\pi}{4} \] \[ \frac{\sin\theta + \cos\theta}{\sqrt{2}} = \sin\left(\theta + \frac{\pi}{4}\right) \] So: \[ E = \tan \left( \sin^{-1}\left(\sin\left(\theta + \frac{\pi}{4}\right)\right) - \theta \right) \]
Step 5: Range check: \[ \theta \in \left[0, \frac{\pi}{6}\right] \Rightarrow \theta + \frac{\pi}{4} \in \left[\frac{\pi}{4}, \frac{5\pi}{12}\right] \] Since this lies within $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$: \[ \sin^{-1}(\sin A) = A \] \[ \Rightarrow \sin^{-1}\left(\sin\left(\theta + \frac{\pi}{4}\right)\right) = \theta + \frac{\pi}{4} \]
Step 6: Simplify: \[ E = \tan \left( \theta + \frac{\pi}{4} - \theta \right) = \tan\left(\frac{\pi}{4}\right) \]
Step 7: Final value: \[ E = 1 \]
Final Answer:
\[ \boxed{1} \]