The problem asks to find the value of $\tan \alpha$, given that:

$\cos \alpha = -\frac{\sqrt{5}}{3} \quad \text{and} \quad \frac{\pi}{2} < \alpha < \pi$

We are given that the angle $\alpha$ is in the second quadrant ($\frac{\pi}{2} < \alpha < \pi$), where $\cos \alpha$ is negative and $\sin \alpha$ is positive.

Step 1: Find $\sin \alpha$ using the Pythagorean identity:

$\sin^2 \alpha + \cos^2 \alpha = 1$

Substitute $\cos \alpha = -\frac{\sqrt{5}}{3}$:

$\sin^2 \alpha + \left(-\frac{\sqrt{5}}{3}\right)^2 = 1$ $\sin^2 \alpha + \frac{5}{9} = 1$ $\sin^2 \alpha = 1 - \frac{5}{9} = \frac{4}{9}$ $\sin \alpha = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Since $\alpha$ is in the second quadrant, $\sin \alpha$ is positive, so $\sin \alpha = \frac{2}{3}$.

Step 2: Calculate $\tan \alpha$

We know that: $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$

Substitute the values of $\sin \alpha = \frac{2}{3}$ and $\cos \alpha = -\frac{\sqrt{5}}{3}$:

$\tan \alpha = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}$

Thus, the value of $\tan \alpha$ is $-\frac{2\sqrt{5}}{5}$.

Answer:

The correct answer is B) $-\frac{2\sqrt{5}}{5}$.

Would you like further details or have any questions? Here are five related questions you could explore:

1. How is the Pythagorean identity derived?
2. What are the signs of trigonometric functions in different quadrants?
3. How do we rationalize the denominator when simplifying fractions?
4. Why is $\sin \alpha$ positive in the second quadrant?
5. How can we find the angle $\alpha$ given $\cos \alpha$?

Tip: For trigonometric problems, always check which quadrant the angle lies in to determine the signs of the sine, cosine, and tangent functions.