Question

sinx/cos3x+sin3x/cos9x+sin9x/cos27x=1/2(tan27x-tanx)

Answer 1

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)