Question
Find the value of:
\[ \tan^{-1}\left(\tan \frac{15\pi}{4}\right) \]
Solution
First, reduce the angle using periodicity of tangent:
\[ \tan(\theta + \pi) = \tan \theta \]
\[ \frac{15\pi}{4} = 3\pi + \frac{3\pi}{4} \]
\[ \tan \frac{15\pi}{4} = \tan \frac{3\pi}{4} \]
\[ \tan \frac{3\pi}{4} = -1 \]
Now evaluate:
\[ \tan^{-1}(-1) \]
The principal value range of \( \tan^{-1}x \) is:
\[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]
Thus,
\[ \tan^{-1}(-1) = -\frac{\pi}{4} \]
Final Answer:
\[ \boxed{-\frac{\pi}{4}} \]
Key Concept
Always reduce the angle first and then apply inverse tangent within its principal range.