1. **State the problem:** Given that $\sin \alpha = \frac{3}{5}$ and $\frac{\pi}{2} < \alpha < \pi$, find the exact values of $\cos \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$. 2. **Recall the formulas:** For any angle $\theta$, the half-angle formulas are: $$\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$ $$\tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{\sin \theta}{1 + \cos \theta}$$ The sign depends on the quadrant of $\frac{\alpha}{2}$. Since $\frac{\pi}{2} < \alpha < \pi$, then $\frac{\pi}{4} < \frac{\alpha}{2} < \frac{\pi}{2}$, so $\cos \frac{\alpha}{2} > 0$ and $\tan \frac{\alpha}{2} > 0$. 3. **Find $\cos \alpha$:** Use the Pythagorean identity $\sin^2 \alpha + \cos^2 \alpha = 1$. $$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$$ We take the negative root because $\alpha$ is in the second quadrant where cosine is negative. 4. **Calculate $\cos \frac{\alpha}{2}$:** $$\cos \frac{\alpha}{2} = +\sqrt{\frac{1 + \cos \alpha}{2}} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{\frac{1}{5}}{2}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$$ 5. **Calculate $\tan \frac{\alpha}{2}$:** Using the formula $\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}$: $$\tan \frac{\alpha}{2} = \frac{\frac{3}{5}}{1 - \frac{4}{5}} = \frac{\frac{3}{5}}{\frac{1}{5}} = 3$$ **Final answers:** $$\cos \frac{\alpha}{2} = \frac{\sqrt{10}}{10}$$ $$\tan \frac{\alpha}{2} = 3$$