Answer
$$\tan\frac{\theta}{2}=3$$
Work Step by Step
$$\sin\theta=\frac{3}{5}\hspace{1.5cm}90^\circ\lt \theta\lt180^\circ\hspace{1.5cm}\tan\frac{\theta}{2}=?$$
Apply the half-angle identity for tangent
$$\tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin\theta}$$
However, we don't have the value of $\cos\theta$ now. So we need to find $\cos\theta$.
1) Find $\cos\theta$
- Pythagorean Identities: $$\cos^2\theta=1-\sin^2\theta=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$$
$$\cos\theta=\pm\frac{4}{5}$$
$90^\circ\lt\theta\lt180^\circ$ denotes that the position of $\theta$ is in quadrant II, where cosines are negative. Thus, $\cos\theta\lt0$.
$$\cos\theta=-\frac{4}{5}$$
2) Find $\tan\frac{\theta}{2}$
Now we can find $\tan\frac{\theta}{2}$ according to the mentioned identity.
$$\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}$$
$$\tan\frac{\theta}{2}=\frac{1-\Big(-\frac{4}{5}\Big)}{\frac{3}{5}}$$
$$\tan\frac{\theta}{2}=\frac{1+\frac{4}{5}}{\frac{3}{5}}$$
$$\tan\frac{\theta}{2}=\frac{\frac{9}{5}}{\frac{3}{5}}$$
$$\tan\frac{\theta}{2}=\frac{9}{3}=3$$