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

A
$a = b$
B
$a = 2 b$
C
$b = 2 a$
D
$a + b = 2$

Solution:

We have,

$a = \tan 27 θ - \tan θ$

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

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

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

Solution:

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