Question

simplify.
$\frac{\tan 5a-\tan 3a}{\tan 5a+\tan 3a}=\frac{\sin 2a}{\sin 8a}$

Answer 1

$\frac{\tan 5a-\tan 3a}{\tan 5a+\tan 3a}$

$=\frac{\frac{\sin 5a}{\cos 5a}-\frac{\sin 3a}{\cos 3a}}{\frac{\sin 5a}{\cos 5a}+\frac{\sin 3a}{\cos 3a}}$

$=\frac{\frac{\sin 5a\cos 3a-\cos 5a\sin 3a}{\cos 5a\cos 3a}}{\frac{\sin 5a\cos 3a+\cos 5a\sin 3a}{\cos 5a\cos 3a}}$

$=\frac{\sin (5a-3a)}{\sin (5a+3a)}$

$=\frac{\sin 2a}{\sin 8a}$