$\frac{\sin \alpha sin11\alpha + sin3\alpha sin7\alpha }{\sin \alpha cos11\alpha + sin3\alpha cos7\alpha }$= $tan8\alpha$
$\frac{\sin \alpha sin11\alpha + sin3\alpha sin7\alpha }{\sin \alpha cos11\alpha + sin3\alpha cos7\alpha }$= $tan8\alpha$ Solution: $\frac{\sin \alpha sin11\alpha + sin3\alpha …