Answer
$$\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}=\tan73.5^\circ$$
Work Step by Step
$$\sqrt{\frac{1-\cos147^\circ}{1+\cos147^\circ}}$$
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.
