## Trigonometry (11th Edition) Clone

$$\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}=\tan73.5^\circ$$
$$\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}$$ 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 $A=147^\circ$: $$\pm\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}=\tan\frac{147^\circ}{2}$$ $$\pm\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}=\tan73.5^\circ$$ Angle $73.5^\circ$ is in quadrant I, where tangent is positive. Therefore, $\tan73.5^\circ\gt0$, and we need to pick the positive square root. $$\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}=\tan73.5^\circ$$ This is the single trigonometric function we need to find.