Question

Resolve that cos3A-cos5A by sin3A+sin5A = tanA

Answer 1

$\frac{2sin(\frac{5A−3A}{2})cos(\frac{5A+3A}{2})}{2cos(\frac{5A+3A}{2})cos(\frac{5A−3A}{2})}$

$\frac{sinA}{cosA}$

$tanA$