Problem
Simplify: \( \tan^{-1}\left(\frac{\sqrt{a – x}}{\sqrt{a + x}}\right), \quad -a < x < a \)
Solution (Substitution Method)
Let:
\[ x = a \cos \theta \]
Then,
\[ a – x = a(1 – \cos \theta), \quad a + x = a(1 + \cos \theta) \]
So,
\[ \frac{\sqrt{a – x}}{\sqrt{a + x}} = \sqrt{\frac{1 – \cos \theta}{1 + \cos \theta}} \]
Using identity:
\[ \frac{1 – \cos \theta}{1 + \cos \theta} = \tan^2 \frac{\theta}{2} \]
Thus,
\[ \frac{\sqrt{a – x}}{\sqrt{a + x}} = \tan \frac{\theta}{2} \]
Hence,
\[ \tan^{-1}\left(\frac{\sqrt{a – x}}{\sqrt{a + x}}\right) = \tan^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} \]
Since \( x = a \cos \theta \), we get:
\[ \theta = \cos^{-1}\left(\frac{x}{a}\right) \]
Final Answer
\[ \boxed{\frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right)} \]