Question
Express the value of:
\[ \tan^{-1}\left(\frac{1}{x}\right) \quad \text{for } x < 0 \]
in terms of \( \cot^{-1}x \).
Solution
Let
\[ \cot^{-1}x = \theta \]
Then,
\[ x = \cot \theta \]
So,
\[ \frac{1}{x} = \tan \theta \]
Thus,
\[ \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(\tan \theta) \]
Now, since \( x < 0 \Rightarrow \theta = \cot^{-1}x \in \left(\frac{\pi}{2}, \pi\right) \)
But the principal range of \( \tan^{-1}x \) is:
\[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]
So we adjust:
\[ \tan^{-1}(\tan \theta) = \theta – \pi \]
Therefore,
\[ \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}x – \pi \]
Final Answer:
\[ \boxed{ \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}x – \pi \quad (x < 0) } \]
Key Concept
Adjust angles based on principal value ranges when converting between inverse trigonometric functions.