Given:
[tex]\sin \frac{\alpha}{2}=\frac{15}{17},90^{\circ}<\alpha<180^{\circ}[/tex]Now,
[tex]\begin{gathered} 90^{\circ}<\alpha<180^{\circ} \\ \text{replace }\alpha\text{ by }\frac{\alpha}{2} \\ \frac{90^{\circ}}{2}<\frac{\alpha}{2}<\frac{180^{\circ}}{2} \\ 45^{\circ}<\frac{\alpha}{2}<90^{\circ} \\ So,\text{ }\frac{\alpha}{2}\text{ lies in first quadrant} \\ \Rightarrow\sin \frac{\alpha}{2},\text{ cos}\frac{\alpha}{2}\text{ and tan}\frac{\alpha}{2}\text{ are positive.} \end{gathered}[/tex]As we know that,
[tex]\begin{gathered} \sin 2\alpha=2\sin \alpha\cos \alpha \\ \text{ replace }\alpha\text{ with }\frac{\alpha}{2} \\ \sin 2(\frac{\alpha}{2})=2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2} \\ \sin \alpha=2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2} \\ 2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}=\frac{15}{17} \\ \sin \frac{\alpha}{2}\cos \frac{\alpha}{2}=\frac{15}{34}\ldots\ldots\ldots.\ldots(1) \end{gathered}[/tex]Now, find the cosine function,
[tex]\begin{gathered} \sin ^2\alpha+\cos ^2\alpha=1 \\ (\frac{15}{17})^2+\cos ^2\alpha=1 \\ \text{cos}^2\alpha=1-\frac{225}{289} \\ \cos ^2\alpha=\frac{64}{289} \\ \cos \alpha=\pm\frac{8}{17} \\ \Rightarrow\cos \alpha=-\frac{8}{17}\ldots.(90^{\circ}\text{<}\alpha<180^{\circ}) \end{gathered}[/tex]Using the identity,
[tex]\begin{gathered} \cos \alpha=2\cos ^2\frac{\alpha}{2}-1 \\ -\frac{8}{17}=2\cos ^2\frac{\alpha}{2}-1 \\ 2\cos ^2\frac{\alpha}{2}=1-\frac{8}{17} \\ \cos ^2\frac{\alpha}{2}=\frac{9}{17\times2} \\ \cos \frac{\alpha}{2}=\pm\sqrt[]{\frac{9}{34}} \\ \Rightarrow\cos \frac{\alpha}{2}=\frac{3}{\sqrt[]{34}}\ldots..(45^{\circ}<\frac{\alpha}{2}<90^{\circ}) \end{gathered}[/tex]From equation (1),
[tex]\begin{gathered} \sin \frac{\alpha}{2}\cos \frac{\alpha}{2}=\frac{15}{34} \\ \sin \frac{\alpha}{2}\times\frac{3}{\sqrt[]{34}}=\frac{15}{34} \\ \sin \frac{\alpha}{2}=\frac{15}{34}\times\frac{\sqrt[]{34}}{3} \\ \sin \frac{\alpha}{2}=\frac{5\sqrt[]{34}}{34} \end{gathered}[/tex]And,
[tex]\tan \frac{\alpha}{2}=\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}=\frac{\frac{5\sqrt[]{34}}{34}}{\frac{3}{\sqrt[]{34}}}=\frac{5\sqrt[]{34}}{34}\times\frac{\sqrt[]{34}}{3}=\frac{5}{3}[/tex]Answer:
[tex]\begin{gathered} \sin \frac{\alpha}{2}=\frac{5\sqrt[]{34}}{34} \\ \tan \frac{\alpha}{2}=\frac{5}{3} \end{gathered}[/tex]