Prove that

(sinA/cos3A) +(sin3A/cos9A) + (sin9A/cos27A) = 1/2(tan27A - tanA)

1. Multiply Nr and Dr by (2 Cos A)

sin A / cos 3A = 2 Sin A Cos A / 2 Cos A Cos 3A = sin 2A / 2 cos A cos 3A = Sin (3A - A) / 2 Cos A Cos 2A

= Sin 3A Cos A - Cos 3A Sin A / 2 Cos A Cos 3A =  1/2 (tan 3A - tan A)

2. Similarly 2nd and 3rd terms are 

1/2 (tan 9A - tan 3A) 

and 

1/2 (tan 27A - tan 9A)

3. Add the three terms. They form a telescopic series. 

Sum = 1/2 (tan 27A - tan A) = RHS