Question

Prove the following:

${\displaystyle \frac{\tan 5\text{A}-\tan 3\text{A}}{\tan 5\text{A}+\tan 3\text{A}}=\frac{\sin 2\text{A}}{\sin 8\text{A}}}$

Answer 1

L.H.S. = ${\displaystyle \frac{\tan 5\text{A}-\tan 3\text{A}}{\tan 5\text{A}+\tan 3\text{A}}}$

= ${\displaystyle \frac{\frac{\sin 5\text{A}}{\cos 5\text{A}}-\frac{\sin 3\text{A}}{\cos 3\text{A}}}{\frac{\sin 5\text{A}}{\cos 5\text{A}}+\frac{\sin 3\text{A}}{\cos 3\text{A}}}}$

= ${\displaystyle \frac{\left(\frac{\sin 5\text{A}\cos 3\text{A}-\sin 3\text{A}\cos 5\text{A}}{\cos 5\text{A}\cos 3\text{A}}\right)}{\left(\frac{\sin 5\text{A}\cos 3\text{A}+\sin 3\text{A}\cos 5\text{A}}{\cos 5\text{A}\cos 3\text{A}}\right)}}$

= ${\displaystyle \frac{\sin 5\text{A}\cos 3\text{A}-\cos 5\text{A}\sin 3\text{A}}{\sin 5\text{A}\cos 3\text{A}+\cos 5\text{A}\sin 3\text{A}}}$

= ${\displaystyle \frac{\sin (5\text{A}-3\text{A})}{\sin (5\text{A}+3\text{A})}}$

= ${\displaystyle \frac{\sin 2\text{A}}{\sin 8\text{A}}}$

= R.H.S.