Question
If
\[ \cos\left(\tan^{-1}x + \cot^{-1}\sqrt{3}\right) = 0 \]
Find \( x \).
Solution
We know:
\[ \cot^{-1}\sqrt{3} = \frac{\pi}{6} \]
So the equation becomes:
\[ \cos\left(\tan^{-1}x + \frac{\pi}{6}\right) = 0 \]
Cosine is zero when:
\[ \theta = \frac{\pi}{2} \quad \text{(within principal consideration)} \]
Thus,
\[ \tan^{-1}x + \frac{\pi}{6} = \frac{\pi}{2} \]
\[ \tan^{-1}x = \frac{\pi}{3} \]
But the range of \( \tan^{-1}x \) is:
\[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]
Since \( \frac{\pi}{3} \) lies in this range,
\[ x = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
Final Answer:
\[ \boxed{\sqrt{3}} \]
Key Concept
Always check the principal value range of inverse trigonometric functions before concluding.