If k1=tan27x−tanx and k2=sinxcos3x+sin3xcos9x+sin9xcos27x then

sinxcos3x=sin2x2cosxcos3x=sin3x−x2cosxcos3x=sin3xcosx−cos3xsinx2cosxcos3x ∴sinxsin3x=12tan3x−tanx................1 sin3xsin9x=12tan9x−tan3x................2 sin9xsin12x=12tan27x−tan9x................3 adding the three equations (1)+(2)+(3)⇒k2=12(tan⁡27x−tan⁡x) ∴ k2=12k1

If k1=tan27x−tanx and k2=sinxcos3x+sin3xcos9x+sin9xcos27x then

sinxcos3x=sin2x2cosxcos3x=sin3x−x2cosxcos3x=sin3xcosx−cos3xsinx2cosxcos3x ∴sinxsin3x=12tan3x−tanx................1 sin3xsin9x=12tan9x−tan3x................2 sin9xsin12x=12tan27x−tan9x................3 adding the three equations (1)+(2)+(3)⇒k2=12(tan⁡27x−tan⁡x) ∴ k2=12k1

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If $k_{1} = \tan 27 x - tanx$ and $k_{2} = \frac{sinx}{\cos 3 x} + \frac{\sin 3 x}{\cos 9 x} + \frac{\sin 9 x}{\cos 27 x}$ then

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$3 k_{1} = k_{2}$

$2 k_{1} = 3 k_{2}$

$k_{1} = 2 k_{2}$

$k_{1} = k_{2}$

answer is B.

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Detailed Solution

$\frac{sinx}{\cos 3 x} = \frac{\sin 2 x}{2 cosxcos 3 x} = \frac{\sin (3 x - x)}{2 cosxcos 3 x} = \frac{\sin 3 xcosx - \cos 3 xsinx}{2 cosxcos 3 x} ∴ \frac{sinx}{\sin 3 x} = \frac{1}{2} (\tan 3 x - tanx) . ............... (1) \frac{\sin 3 x}{\sin 9 x} = \frac{1}{2} (\tan 9 x - \tan 3 x) . ............... (2) \frac{\sin 9 x}{\sin 12 x} = \frac{1}{2} (\tan 27 x - \tan 9 x) . ............... (3) adding the three equations (1) + (2) + (3) \Rightarrow k_{2} = \frac{1}{2} (\tan 27 x - \tan x) ∴ k_{2} = \frac{1}{2} k 1$