Show that cos3xsinx+cos9xsin 3x+cos27xsin9x=21[tan27x−tanx].

cos3x sinx +cos9x sin3x +cos27x sin9x multiplying and dividing by 2 =21[cos3x 2sinx +cos9x 2sin3x +cos27x 2sin9x ]multiplying and dividing by cosx,cos9x,cos27 respectively and simplifying =21[cos3xcosx sin2x +cos9xcos3x sin6x +cos27x cos9xsin18x ]Use, sin(A−B)=sinAcosB−cosAsinB =21[cos3xcosx sin3xcosx−cos3xsinx ]+21[cos27xcos9x sin9xcos3x−cos9xsin3x ] +21[cos27xcos9x sin27xcos9x−cos27xsin9x ] =21[tan27x−tan9x+tan9x−tan3x+tan3x−tanx ]=21[tan27x−tanx ]

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Show that { x}{ 3x}+{ 3x}{ 9x}+{ 9x}{ 27x}={1}{2} [ 27x- x

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Show that $\frac{sin x}{cos 3 x} + \frac{sin 3 x}{cos 9 x} + \frac{sin 9 x}{cos 27 x} = \frac{1}{2} [tan 27 x - tan x]$ .

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\frac{sin x}{cos 3 x} + \frac{sin 3 x}{cos 9 x} + \frac{sin 9 x}{cos 27 x} = \frac{1}{2} [tan 27 x - tan x]

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Text solutionVerified

\frac{sin x}{cos 3 x} + \frac{sin 3 x}{cos 9 x} + \frac{sin 9 x}{cos 27 x}

multiplying and dividing by 2

= \frac{1}{2} [\frac{2 sin x}{cos 3 x} + \frac{2 sin 3 x}{cos 9 x} + \frac{2 sin 9 x}{cos 27 x}]

multiplying and dividing by

cos x, cos 9 x, cos 27

respectively and simplifying

= \frac{1}{2} [\frac{sin 2 x}{cos 3 x cos x} + \frac{sin 6 x}{cos 9 x cos 3 x} + \frac{sin 18 x}{cos 27 x cos 9 x}]

Use,

sin (A - B) = sin A cos B - cos A sin B

= \frac{1}{2} [\frac{sin 3 x cos x - cos 3 x sin x}{cos 3 x cos x}] + \frac{1}{2} [\frac{sin 9 x cos 3 x - cos 9 x sin 3 x}{cos 27 x cos 9 x}]

+ \frac{1}{2} [\frac{sin 27 x cos 9 x - cos 27 x sin 9 x}{cos 27 x cos 9 x}]

= \frac{1}{2} [tan 27 x - tan 9 x + tan 9 x - tan 3 x + tan 3 x - tan x] = \frac{1}{2} [tan 27 x - tan x]

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Question Text	Show that $\frac{sin x}{cos 3 x} + \frac{sin 3 x}{cos 9 x} + \frac{sin 9 x}{cos 27 x} = \frac{1}{2} [tan 27 x - tan x]$ .
Answer Type	Text solution:1
Upvotes	150

Show that cos3xsinx​+cos9xsin 3x​+cos27xsin9x​=21​[tan27x−tanx].

Text solutionVerified

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Show that $\frac{sin x}{cos 3 x} + \frac{sin 3 x}{cos 9 x} + \frac{sin 9 x}{cos 27 x} = \frac{1}{2} [tan 27 x - tan x]$ .