## Trigonometry (11th Edition) Clone

$$\sqrt{\frac{1+\cos165^\circ}{1-\cos165^\circ}}=\cot82.5^\circ$$
$$X=\sqrt{\frac{1+\cos165^\circ}{1-\cos165^\circ}}$$ 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.