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,
\[ x = \tan \theta \]
Now,
\[ \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{\tan\theta}\right) = \tan^{-1}(\cot\theta) \]
We know that:
\[ \cot\theta = \tan\left(\frac{\pi}{2} – \theta\right) \]
So,
\[ \tan^{-1}(\cot\theta) = \frac{\pi}{2} – \theta \quad \text{(since } x > 0 \Rightarrow \theta > 0\text{)} \]
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 positive values of \( x \), the angles are complementary, hence their sum is \( \frac{\pi}{2} \).