#### Answer

$$\pm\sqrt{\frac{1-\cos8\theta}{1+\cos8\theta}}=\tan4\theta$$

#### Work Step by Step

$$\pm\sqrt{\frac{1-\cos8\theta}{1+\cos8\theta}}$$
Here we notice at the start of the expression, there is the sign $\pm$, which means we do not have to decide whether to take positive or negative square root.
From the half-angle identity for tangents:
$$\pm\sqrt{\frac{1-\cos A}{1+\cos A}}=\tan\frac{A}{2}$$
We can apply the identity to the given expression with $A=8\theta$.
$$\pm\sqrt{\frac{1-\cos8\theta}{1+\cos8\theta}}=\tan\frac{8\theta}{2}$$
$$\pm\sqrt{\frac{1-\cos8\theta}{1+\cos8\theta}}=\tan4\theta$$