Question
If
\[ \tan^{-1}\left(\frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right) = \alpha \]
Find \( x^2 \).
Solution
Let
\[ \tan^{-1}(A) = \alpha \Rightarrow A = \tan \alpha \]
So,
\[ \tan \alpha = \frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} \]
This is a standard identity:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{1 – \cos \theta}{\sin \theta} \quad \text{or equivalently} \]
Recognize that:
\[ \frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} = \tan\left(\frac{1}{2}\sin^{-1}x\right) \]
Thus,
\[ \alpha = \frac{1}{2}\sin^{-1}x \]
\[ \sin^{-1}x = 2\alpha \Rightarrow x = \sin 2\alpha \]
Therefore,
\[ x^2 = \sin^2 2\alpha \]
Final Answer:
\[ \boxed{x^2 = \sin^2 2\alpha} \]
Key Concept
Recognize standard transformation forms involving square roots and inverse trigonometric identities.