If k1=tan⁡27θ−tan⁡θ and k2=sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ then

We can write k1=tan⁡27θ−tan⁡9θ+tan⁡9θ−tan⁡3θ+tan⁡3θ−tan⁡θBut tan⁡3θtan⁡θ−sin⁡3θcos⁡θ−cos⁡3θsin⁡θcos⁡3θcos⁡θ=sin⁡2θcos⁡3θcos⁡θ=2sin⁡θcos⁡3θk1=2sin⁡9θcos⁡27θ+sin⁡3θcos⁡9θ+sin⁡θcos⁡3θ=2k2.

If k1=tan⁡27θ−tan⁡θ and k2=sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ then

We can write k1=tan⁡27θ−tan⁡9θ+tan⁡9θ−tan⁡3θ+tan⁡3θ−tan⁡θBut tan⁡3θtan⁡θ−sin⁡3θcos⁡θ−cos⁡3θsin⁡θcos⁡3θcos⁡θ=sin⁡2θcos⁡3θcos⁡θ=2sin⁡θcos⁡3θk1=2sin⁡9θcos⁡27θ+sin⁡3θcos⁡9θ+sin⁡θcos⁡3θ=2k2.

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

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

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

$k_{1} = k_{2}$

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

answer is B.

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Detailed Solution

We can write

$k_{1} = \tan 27 θ - \tan 9 θ + \tan 9 θ - \tan 3 θ + \tan 3 θ -$ $\tan θ$

But $\begin{aligned} \tan 3 θ \tan θ - \frac{\sin 3 θ \cos θ - \cos 3 θ \sin θ}{\cos 3 θ \cos θ} \\ = \frac{\sin 2 θ}{\cos 3 θ \cos θ} = \frac{2 \sin θ}{\cos 3 θ} \\ k_{1} = 2 [\frac{\sin 9 θ}{\cos 27 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin θ}{\cos 3 θ}] = 2 k_{2} . \end{aligned}$