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# (sin x / cos 3x)+(sin 3x /cos 9x)+(sin 9x/cos 27x)=(1/2)*(tan 27x- tan x) suryakanth AskiitiansExpert-IITB
105 Points
10 years ago

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.

3 years ago
16yrfhu5tthu ucuc7f7f7f77c8g9vb0npnovyxuv8c6uc8g9g 8g8gg87ff78gv99v9b0h0jb099vc87cc78c9vn00jguuc8v8v8v8v8v8vv99h9b9b0j0jn09vv98cc77c7c7c7c7c Kushagra Madhukar
10 months ago
Dear student,

Let us 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 adding equation (1)+(2)+(3) we get
(sinx / cos3x) + (sin3x / cos9x) + (sin9x / cos27x)
= (1/2) * (tan3x - tanx + tan9x - tan3x + tan27x - tan9x)
= (1/2) * (tan27x - tanx)
= RHS
LHS = RHS
Hence proved.

Thanks and regards,
Kushagra