Question

Help me solve this question

Answer 1

Consider the first term on the left side.

$\displaystyle \frac{2\sin{A}}{\cos{3A}}\\\\=\frac{2\sin{A}\cos{A}}{\cos{3A}\cos{A}}\qquad\text{multiply by cos(A)/cos(A)}\\\\=\frac{\sin{2A}}{\cos{3A}\cos{A}}\qquad\text{double angle formula for sin}\\\\=\frac{\sin{(2A-A)}}{\cos{3A}\cos{A}}=\frac{\sin{3A}\cos{A}-\cos{3A}\sin{A}}{\cos{3A}\cos{A}}\\\\=\tan{3A}-\tan{A}$

Then the sum of the three terms on the left is

$\left(\tan{3A}-\tan{A}\right)+\left(\tan{9A}-\tan{3A}\right)+\left(\tan{27A}-\tan{9A}\right)\\\\=\tan{27A}-\tan{A}\qquad\text{Q.E.D.}$