Question
Find the value of:
\[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) \] given that \( x < 0 \).
Solution
Let
\[ \tan^{-1}x = \theta \]
Then,
\[ x = \tan \theta \]
Since \( x < 0 \Rightarrow \theta < 0 \).
Now,
\[ \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(\cot\theta) \]
We know:
\[ \cot\theta = \tan\left(\frac{\pi}{2} – \theta\right) \]
But since \( \theta < 0 \), the angle \( \frac{\pi}{2} - \theta > \frac{\pi}{2} \), which lies outside the principal range of \( \tan^{-1} \).
So, we adjust:
\[ \tan^{-1}(\cot\theta) = \frac{\pi}{2} – \theta – \pi = -\frac{\pi}{2} – \theta \]
Therefore,
\[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \theta + \left(-\frac{\pi}{2} – \theta\right) \]
\[ = -\frac{\pi}{2} \]
Final Answer:
\[ \boxed{ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2} \quad \text{for } x < 0 } \]
Key Concept
For negative values of \( x \), the sum of inverse tangents becomes \( -\frac{\pi}{2} \) due to principal value adjustment.