Answer
$$\cot\theta=\frac{\sqrt5}{5}$$
Work Step by Step
$$\tan\theta=-\frac{\sqrt5}{2}\hspace{1.5cm}90^\circ\lt\theta\lt180^\circ\hspace{1.5cm}\cot\frac{\theta}{2}=?$$
To find $\cot\frac{\theta}{2}$, first we need to find $\tan\frac{\theta}{2}$, which involves using half-angle identity for tangents:
$$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$
However, both $\sin\theta$ and $\cos\theta$ are not known now, meaning we need to find them.
1) Find $\sin\theta$ and $\cos\theta$
The question states that $90^\circ\lt\theta\lt180^\circ$, meaning the angle $\theta$ lies in quadrant II, where sines are positive but cosines are negative.
Thus, $\sin\theta\gt0$ and $\cos\theta\lt0$.
- Pythagorean Identities:
$$\sec^2\theta=1+\tan^2\theta=1+\Big(-\frac{\sqrt5}{2}\Big)^2=1+\frac{5}{4}=\frac{9}{4}$$
$$\sec\theta=\pm\frac{3}{2}$$
- Reciprocal Identities:
$$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\pm\frac{3}{2}}=\pm\frac{2}{3}$$
But $\cos\theta\lt0$, so $$\cos\theta=-\frac{2}{3}$$
- Quotient Identities:
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
$$\sin\theta=\tan\theta\times\cos\theta=-\frac{\sqrt5}{2}\times\Big(-\frac{2}{3}\Big)=\frac{\sqrt5}{3}$$
Therefore, $$\sin\theta=\frac{\sqrt5}{3}\hspace{2cm}\cos\theta=-\frac{2}{3}$$
2) Find $\tan\frac{\theta}{2}$
Now we can apply the identity to find $\tan\frac{\theta}{2}$
$$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$
$$\tan\frac{\theta}{2}=\frac{\frac{\sqrt5}{3}}{1-\frac{2}{3}}$$
$$\tan\frac{\theta}{2}=\frac{\frac{\sqrt5}{3}}{\frac{1}{3}}$$
$$\tan\frac{\theta}{2}=\frac{\sqrt5}{1}=\sqrt5$$
3) Find $\cot\frac{\theta}{2}$
- Reciprocal Identities:
$$\cot\frac{\theta}{2}=\frac{1}{\tan\frac{\theta}{2}}=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}$$
