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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
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