Problem
Simplify: \( \tan^{-1}\left(\frac{\sqrt{1 + x^2} – 1}{x}\right), \quad x \ne 0 \)
Solution (Substitution Method)
Let:
\[ x = \tan \theta \]
Then,
\[ \sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta \]
So the expression becomes:
\[ \tan^{-1}\left(\frac{\sec \theta – 1}{\tan \theta}\right) \]
Simplify:
\[ \frac{\sec \theta – 1}{\tan \theta} = \frac{\frac{1}{\cos \theta} – 1}{\frac{\sin \theta}{\cos \theta}} = \frac{1 – \cos \theta}{\sin \theta} \]
Using identity:
\[ \frac{1 – \cos \theta}{\sin \theta} = \tan \frac{\theta}{2} \]
Thus,
\[ \tan^{-1}\left(\frac{\sqrt{1 + x^2} – 1}{x}\right) = \tan^{-1}(\tan \frac{\theta}{2}) = \frac{\theta}{2} \]
Since \( \theta = \tan^{-1} x \), we get:
Final Answer
\[ \boxed{\frac{1}{2}\tan^{-1} x} \]