Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 694: 76

$\displaystyle \tan\frac{5A}{2}$

Comparing with the Half-Angle Identity $\displaystyle \tan\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{1+\cos A}}$, if we replace $A$ with $5A$ in the identity, the RHS equals the given expression. So, the LHS =$ \displaystyle \tan\frac{5A}{2}$