Question
If \( x < 0 \), \( y < 0 \) and \( xy = 1 \), find:
\[ \tan^{-1}x + \tan^{-1}y \]
Solution
We use identity:
\[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x + y}{1 – xy}\right) \]
Given:
\[ xy = 1 \]
So denominator becomes:
\[ 1 – xy = 1 – 1 = 0 \]
This implies the expression tends to:
\[ \tan^{-1}(\infty) = \frac{\pi}{2} \quad \text{or} \quad \tan^{-1}(-\infty) = -\frac{\pi}{2} \]
Now since \( x < 0 \), \( y < 0 \), their sum is negative:
\[ x + y < 0 \]
Thus the expression tends to:
\[ -\frac{\pi}{2} \]
Final Answer:
\[ \boxed{-\frac{\pi}{2}} \]
Key Concept
When denominator becomes zero, check sign of numerator to determine whether result is \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \).