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[Post New]13 Jun 2008 18:57:52 IST
    Subject: Re:Prove that
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hsbhatt (5849)

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Olaaa!! Perrrfect answer. 1101  [1273 rates]
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\text{We have} \ \frac{1}{2} (\tan 3x - \tan x) \\ \\<br/>= \frac{1}{2} \left( \frac{\sin 3x}{\cos 3x} - \frac{\sin x}{\cos x} \right ) \\ \\<br/>= \frac{ \sin 3x \cos x - \cos 3x \sin x}{2 \cos x \cos 3x} \\ \\<br/>= \frac{\sin 2x}{2 \cos x \cos 3x} = \frac{ 2 \sin x \cos x}{2 \cos x \cos 3x} \\ \\<br/>= \frac{\sin x}{\cos 3x} \\ \\<br/>\text{Thus, we can write} \\ \\<br/>\frac{1}{2} (\tan 3x - \tan x) = \frac{\sin x}{\cos 3x} \\ \\<br/>\frac{1}{2} (\tan 9x - \tan 3x) = \frac{\sin 3x}{\cos 9x} \\ \\<br/>\frac{1}{2} (\tan 27x - \tan 9x) = \frac{\sin 9x}{\cos 27x}<br/>\\ \\<br/>\text{Adding the three, we get} \\ \\<br/>\frac{1}{2} (\tan 27x - \tan x) = \frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x}<br/>
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