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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.

Егер cos α = -√5/3, π/2 < α < π болса, онда tg α-ны есептеңіз.

Learn how to calculate tan α when cos α = -√5/3 and π/2 < α < π. This problem uses trigonometric identities like the Pythagorean Identity and covers how to find the tangent of an angle in the second quadrant. Suitable for Grades 10-12.

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