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

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.

```
7 years ago
sky
13 Points
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4 months ago
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