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