If sinθcos3θ+sin3θcos9θ+sin9θcos27θ+sin27θcos81θ=AtanBθ−tanCθ then B−CA= A>0,θ≠2n+1π2,n∈Z

We have $\frac{\sin θ}{\cos 3 θ} = \frac{1}{2} (\frac{\sin 2 θ}{\cos θ \cos 3 θ}) = \frac{1}{2} (\tan 3 θ - \tan θ)$

Similarly $\frac{\sin 3 θ}{\cos 9 θ} = \frac{1}{2} (\tan 9 θ - \tan 3 θ)$

$\frac{\sin 9 θ}{\cos 27 θ} = \frac{1}{2} (\tan 27 θ - \tan 9 θ)$ and $\frac{\sin 27 θ}{\cos 81 θ} = \frac{1}{2} (\tan 81 θ - \tan 27 θ)$

$∴ \frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} + \frac{\sin 27 θ}{\cos 81 θ} = \frac{1}{2} (\tan 81 θ - \tan θ)$

$\Rightarrow \frac{B - C}{A} = \frac{81 + 1}{1 / 2} = 164$

If $\frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} + \frac{\sin 27 θ}{\cos 81 θ} = A (\tan B θ - \tan C θ)$ then $(\frac{B - C}{A}) = (A > 0, θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$