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If $a = \tan 27 θ - \tan θ$ and $b = \frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ},$ then

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a = b

a = 2 b

b = 2 a

a + b = 2

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detailed solution

Correct option is B

We have,

$a = \tan 27 θ - \tan θ$

$\begin{array}{l} = a = \tan 27 θ - \tan 9 θ + \tan 9 θ - \tan 3 θ + \tan 3 θ - \tan θ \\ \Rightarrow a = \frac{\sin 18 θ}{\cos 27 θ \cos 9 θ} + \frac{\sin 6 θ}{\cos 9 θ \cos 3 θ} + \frac{\sin 2 θ}{\cos 3 θ \cos θ} \end{array}$

$\Rightarrow a = 2 (\frac{\sin 9 θ}{\cos 27 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin θ}{\cos 3 θ}) \Rightarrow a = 2 b$

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If a=tan⁡27θ−tan⁡θ and b=sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ, then