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divya vikash Grade: 11
        

 (sin x / cos 3x)+(sin 3x /cos 9x)+(sin 9x/cos 27x)=(1/2)*(tan 27x- tan x)

6 years ago

Answers : (1)

suryakanth AskiitiansExpert-IITB
105 Points
										

Dear Divya,


Lets consider (sinx / cos3x)


(sinx / cos3x)
= (1/2) * [(2 * sinx * cosx) / (cosx * cos3x)]
= (1/2) * [(sin2x) / (cosx * cos3x)]
= (1/2) * [sin(3x - x) / (cosx * cos3x)]
= (1/2) * [(sin3x * cosx - cos3x * sinx) / (cosx * cos3x)]
= (1/2) * [{(sin3x * cosx) / (cosx * cos3x)} - {(cos3x * sinx) / (cosx * cos3x)}]
= (1/2) * [(sin3x/cos3x) - (sinx/cosx)]
= (1/2) * (tan3x - tanx)

so, (sinx / cos3x) = (1/2) * (tan3x - tanx) ---------(1)

Now putting 3A in place of A in equation (1) we get
(sin3x / cos9x) = (1/2) * (tan9x - tan3x) ------------(2)

Again putting 9A in place of A in equation (1) we get
(sin9x / cos27x) = (1/2) * (tan27x - tan9x) ------------(3)

Now doing (1)+(2)+(3) we get
(sinx / cos3x) + (sin3x / cos9x) + (sin9x / cos27x)
= (1/2) * (tan3x - tanx + tan9x - tan3x + tan27x - tan9x)
= (1/2) * (tan27x - tanx)


Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.



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6 years ago
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