Question

If $\tan^{-1} \left[ \frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} \right] = \alpha$, then the value of $\sin 2\alpha$ is

• A. $x^3$
• B. $x$
• C. $x^4$
• D. $x^2$

Hint:

When stuck on picking a substitution for expressions containing both $\sqrt{1+u}$ and $\sqrt{1-u}$, always use $u = \cos\theta$. The number 1 pairs perfectly with cosine to form clean, single-term squares ($2\cos^2\frac{\theta}{2}$ or $2\sin^2\frac{\theta}{2}$) that instantly eliminate square roots!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an inverse trigonometric equation involving radical expressions. We need to find the value of $\sin 2\alpha$ in terms of $x$.

Step 2: Key Formula or Approach:
To simplify the complex fraction inside the inverse tangent, use trigonometric substitution. Let $x^2 = \cos \theta$. This allows us to use the standard half-angle identities to eliminate the radicals: $$1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) \quad \text{and} \quad 1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$$

Step 3: Detailed Explanation:
1. Substitute $x^2 = \cos \theta$ into the expression: $$\tan \alpha = \frac{\sqrt{1 + \cos \theta} - \sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta} + \sqrt{1 - \cos \theta}}$$ 2. Apply the half-angle identities inside the square roots: $$\tan \alpha = \frac{\sqrt{2 \cos^2(\theta/2)} - \sqrt{2 \sin^2(\theta/2)}}{\sqrt{2 \cos^2(\theta/2)} + \sqrt{2 \sin^2(\theta/2)}} = \frac{\sqrt{2}\cos(\theta/2) - \sqrt{2}\sin(\theta/2)}{\sqrt{2}\cos(\theta/2) + \sqrt{2}\sin(\theta/2)}$$ 3. Cancel the $\sqrt{2}$ factor from the numerator and denominator: $$\tan \alpha = \frac{\cos(\theta/2) - \sin(\theta/2)}{\cos(\theta/2) + \sin(\theta/2)}$$ 4. Divide both the numerator and the denominator by $\cos(\theta/2)$ to transform the expression into tangent terms: $$\tan \alpha = \frac{1 - \tan(\theta/2)}{1 + \tan(\theta/2)}$$ 5. Use the tangent subtraction formula, noting that $1 = \tan(\frac{\pi}{4})$: $$\tan \alpha = \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \implies \alpha = \frac{\pi}{4} - \frac{\theta}{2}$$ 6. Multiply the entire equation by 2 to match the target expression: $$2\alpha = \frac{\pi}{2} - \theta$$ 7. Take the sine of both sides and apply the co-function reflection identity ($\sin(\frac{\pi}{2} - \theta) = \cos \theta$): $$\sin 2\alpha = \sin \left( \frac{\pi}{2} - \theta \right) = \cos \theta$$ 8. Since we initially defined $x^2 = \cos \theta$, substitute it back: $$\sin 2\alpha = x^2$$

Step 4: Final Answer:
The value of $\sin 2\alpha$ is $x^2$, which corresponds to option (D).