Problem
Simplify: \( \sin\left(2\tan^{-1}\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right)\right) \)
Solution (Substitution Method)
Let:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right) \]
Then,
\[ \tan \theta = \frac{\sqrt{1-x}}{\sqrt{1+x}} \]
Using identity:
\[ \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \]
Compute:
\[ \tan^2 \theta = \frac{1-x}{1+x} \]
\[ 1 + \tan^2 \theta = \frac{1+x + 1-x}{1+x} = \frac{2}{1+x} \]
Thus,
\[ \sin 2\theta = \frac{2 \cdot \frac{\sqrt{1-x}}{\sqrt{1+x}}}{\frac{2}{1+x}} \]
\[ = \frac{2\sqrt{1-x}}{\sqrt{1+x}} \cdot \frac{1+x}{2} \]
\[ = \sqrt{1-x}\sqrt{1+x} \]
\[ = \sqrt{1 – x^2} \]
Final Answer
\[ \boxed{\sqrt{1 – x^2}} \]