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Q sinx/cos3x + sin3x/cos9x +sin9x/cos27x = ? answer is 1/2(tan27x-tanx) Please give the solution of the question

Q    sinx/cos3x + sin3x/cos9x +sin9x/cos27x = ?   answer is 1/2(tan27x-tanx)
Please give the solution of the question

Grade:11

1 Answers

Latika Leekha
askIITians Faculty 165 Points
8 years ago
Hello student,
sinx/cos3x + sin3x/cos9x + sin9x/cos27x
= \frac{2 sin x cos x}{2 cos 3x cos x} + \frac{2 sin 3x cos 3x}{2 cos 9x cos 3x} + \frac{2 sin 9x cos 9x}{2 cos 27x cos 9x}
= \frac{sin 2x}{2 cos 3x cos x} + \frac{sin 6x}{2 cos 9x cos 3x} + \frac{sin 18x}{2 cos 27x cos 9x}
= \frac{sin (3x-x)}{2 cos 3x cos x} + \frac{sin (9x-3x)}{2 cos 9x cos 3x} + \frac{sin (27x-9x)}{2 cos 27x cos 9x}
= ½ { tan 3x – tan x + tan 9x – tan 3x + tan 27x – tan 9x}
= ½ ( tan 27 x – tan x)

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