Question

Given $0 \le x \le \dfrac{1}{2}$ then the value of $\tan \left[ \sin^{-1} \left( \dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{1-x^2}}{\sqrt{2}} \right) - \sin^{-1} x \right]$ is:

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

Hint:

Whenever you see $\frac{1}{\sqrt{2}}$ paired with $x$ and $\sqrt{1-x^2}$, it is a strong hint to use a compound angle formula involving $\pi/4$ (or $45^\circ$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires simplifying the argument of the tangent function using inverse trigonometric substitutions. We can use the substitution $x = \sin \theta$ to simplify the expression inside the first $\sin^{-1}$ term.

Step 2: Key Formula or Approach:
Use the identity $\sin A \cos B + \cos A \sin B = \sin(A+B)$ and the substitution $x = \sin \theta$.

Step 3: Detailed Explanation:
1. Let $x = \sin \theta$. Then $\sqrt{1-x^2} = \cos \theta$. 2. The expression inside the first $\sin^{-1}$ becomes: \[ \frac{1}{\sqrt{2}}\sin \theta + \frac{1}{\sqrt{2}}\cos \theta \] 3. Since $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, this is: \[ \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \sin\left(\theta + \frac{\pi}{4}\right) \] 4. Substitute this back into the original expression: \[ \tan \left[ \sin^{-1} \left( \sin\left(\theta + \frac{\pi}{4}\right) \right) - \sin^{-1}(\sin \theta) \right] \] 5. Simplify the inverse functions: \[ \tan \left[ \left(\theta + \frac{\pi}{4}\right) - \theta \right] = \tan \frac{\pi}{4} \] 6. Calculate the final value: \[ \tan \frac{\pi}{4} = 1 \]

Step 4: Final Answer
The value of the expression is 1.