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: Name the inside expression.
We have $\tan\alpha = \dfrac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}$, and we want $\sin 2\alpha$. A neat route is to find $\cos 2\alpha$ via $\dfrac{1 - \tan^2\alpha}{1 + \tan^2\alpha}$.
Step 2: Rationalise to simplify $\tan\alpha$.
Multiply top and bottom by $\sqrt{1+x^2} - \sqrt{1-x^2}$. The denominator becomes $(1+x^2)-(1-x^2) = 2x^2$, and the numerator becomes $\left(\sqrt{1+x^2}-\sqrt{1-x^2}\right)^2$.
Step 3: Expand the numerator.
$\left(\sqrt{1+x^2}-\sqrt{1-x^2}\right)^2 = (1+x^2) + (1-x^2) - 2\sqrt{(1+x^2)(1-x^2)} = 2 - 2\sqrt{1-x^4}$. So $\tan\alpha = \dfrac{2 - 2\sqrt{1-x^4}}{2x^2} = \dfrac{1 - \sqrt{1-x^4}}{x^2}$.
Step 4: Switch to a substitution for $\cos 2\alpha$.
Let $x^2 = \cos\theta$. Then $\sqrt{1-x^4} = \sqrt{1-\cos^2\theta} = \sin\theta$, so $\tan\alpha = \dfrac{1 - \sin\theta}{\cos\theta}$. Using $1-\sin\theta$ and $\cos\theta$ in half-angle form, this equals $\tan\!\left(\dfrac{\pi}{4} - \dfrac{\theta}{2}\right)$.
Step 5: Read off $\alpha$ and double it.
Thus $\alpha = \dfrac{\pi}{4} - \dfrac{\theta}{2}$, giving $2\alpha = \dfrac{\pi}{2} - \theta$.
Step 6: Take the sine and restore $x$.
$\sin 2\alpha = \sin\!\left(\dfrac{\pi}{2} - \theta\right) = \cos\theta = x^2$.
\[ \boxed{\sin 2\alpha = x^2} \]