## Precalculus (6th Edition)

$\tan 4\theta$
Comparing with the Half-Angle Identity $\displaystyle \tan\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{1+\cos A}}$, if we define $A=8\theta$, the RHS equals the given expression. So, the LHS =$\tan 4\theta$