Answer
$$\sqrt{\frac{1+\cos165^\circ}{1-\cos165^\circ}}=\cot82.5^\circ$$
Work Step by Step
$$X=\sqrt{\frac{1+\cos165^\circ}{1-\cos165^\circ}}$$
Recall the half-angle identity for tangent, which states
$$\pm\sqrt{\frac{1-\cos A}{1+\cos A}}=\tan\frac{A}{2}$$
Apply the identity for $A=165^\circ$:
$$\pm\sqrt{\frac{1-\cos165^\circ}{1+\cos165^\circ}}=\tan\frac{165^\circ}{2}$$
$$\pm\sqrt{\frac{1-\cos165^\circ}{1+\cos165^\circ}}=\tan82.5^\circ$$
Angle $82.5^\circ$ is in quadrant I, where tangent is positive. Therefore, $\tan82.5^\circ\gt0$, and we need to pick the positive square root.
$$\sqrt{\frac{1-\cos165^\circ}{1+\cos165^\circ}}=\tan82.5^\circ$$
However, the sign of $X$ is still different from the equation just found, so we need to rewrite a little bit.
$$\frac{\sqrt{1-\cos165^\circ}}{\sqrt{1+\cos165^\circ}}=\tan82.5^\circ$$
Thus,
$$\frac{\sqrt{1+\cos165^\circ}}{\sqrt{1-\cos165^\circ}}=\frac{1}{\tan82.5^\circ}$$
$$\sqrt{\frac{1+\cos165^\circ}{1-\cos165^\circ}}=\frac{1}{\tan82.5^\circ}$$
$$X=\frac{1}{\tan82.5^\circ}$$
However, as $\frac{1}{\tan\theta}=\cot\theta$,
$$X=\cot82.5^\circ$$
Therefore, $$\sqrt{\frac{1+\cos165^\circ}{1-\cos165^\circ}}=\cot82.5^\circ$$
This is the single trigonometric function we need to find.
