Problem
Simplify: \( \tan^{-1}(\sqrt{1 + x^2} – x), \quad x \in \mathbb{R} \)
Solution (Substitution Method)
Let:
\[ x = \tan \theta \]
Then,
\[ \sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta \]
So the expression becomes:
\[ \tan^{-1}(\sec \theta – \tan \theta) \]
Now use identity:
\[ \sec \theta – \tan \theta = \frac{1}{\sec \theta + \tan \theta} \]
Also,
\[ \sec \theta + \tan \theta = \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) \]
Thus,
\[ \sec \theta – \tan \theta = \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) \]
Hence,
\[ \tan^{-1}(\sec \theta – \tan \theta) = \tan^{-1}\left[\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right] \]
Using identity:
\[ \tan^{-1}(\cot \alpha) = \frac{\pi}{2} – \alpha \]
So,
\[ = \frac{\pi}{2} – \left(\frac{\pi}{4} + \frac{\theta}{2}\right) \]
\[ = \frac{\pi}{4} – \frac{\theta}{2} \]
Since \( \theta = \tan^{-1} x \), we get:
Final Answer
\[ \boxed{\frac{\pi}{4} – \frac{1}{2}\tan^{-1} x} \]