To solve $\tan^{-1} \left( \tan \left( -\frac{8\pi}{9} \right) \right)$, we need to understand how the inverse tangent function works and take note of the range of the principal values for $\tan^{-1}$.

Step-by-Step Solution

1. Understanding the Range of $\tan^{-1}(x)$: The range of $\tan^{-1}(x)$ is $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$. This means that any output of $\tan^{-1}(x)$ must be within this interval.

2. Analyzing $\tan \left( -\frac{8\pi}{9} \right)$: We know that $-\frac{8\pi}{9}$ is outside the range of $\tan^{-1}$. Therefore, we need to find an equivalent angle for $-\frac{8\pi}{9}$ that lies within $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ and has the same tangent value.

3. Finding an Equivalent Angle in the Correct Range: Since $\tan(x)$ has a period of $\pi$, we can add or subtract multiples of $\pi$ to bring the angle into the range $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.

   Adding $\pi$ to $-\frac{8\pi}{9}$: $-\frac{8\pi}{9} + \pi = -\frac{8\pi}{9} + \frac{9\pi}{9} = \frac{\pi}{9}$

4. Conclusion: The angle $\frac{\pi}{9}$ is within the range $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ and has the same tangent as $-\frac{8\pi}{9}$. Therefore: $\tan^{-1} \left( \tan \left( -\frac{8\pi}{9} \right) \right) = \frac{\pi}{9}$

Final Answer:

$\frac{\pi}{9}$

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Related Questions

1. How does the tangent function’s periodicity affect inverse trigonometric calculations?
2. What is the principal range of each trigonometric function and its inverse?
3. Why do we need to adjust angles to be within the principal range of inverse trigonometric functions?
4. How can we find equivalent angles with the same tangent, sine, or cosine values?
5. What would happen if the angle were in degrees instead of radians?

Tip:

For inverse trigonometric functions, always check the principal range to ensure your answer falls within the correct interval.