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

# If $\frac{\mathrm{sin}\theta }{\mathrm{cos}3\theta }+\frac{\mathrm{sin}3\theta }{\mathrm{cos}9\theta }+\frac{\mathrm{sin}9\theta }{\mathrm{cos}27\theta }+\frac{\mathrm{sin}27\theta }{\mathrm{cos}81\theta }=A\left(\mathrm{tan}B\theta -\mathrm{tan}C\theta \right)$  then

1. A
2. B
3. C
4. D

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### Solution:

We have$\frac{\mathrm{sin}\theta }{\mathrm{cos}3\theta }=\frac{1}{2}\left(\frac{\mathrm{sin}2\theta }{\mathrm{cos}\theta \mathrm{cos}3\theta }\right)=\frac{1}{2}\left(\mathrm{tan}3\theta -\mathrm{tan}\theta \right)$

Similarly $\frac{\mathrm{sin}3\theta }{\mathrm{cos}9\theta }=\frac{1}{2}\left(\mathrm{tan}9\theta -\mathrm{tan}3\theta \right)$

$\frac{\mathrm{sin}9\theta }{\mathrm{cos}27\theta }=\frac{1}{2}\left(\mathrm{tan}27\theta -\mathrm{tan}9\theta \right)$ and $\frac{\mathrm{sin}27\theta }{\mathrm{cos}81\theta }=\frac{1}{2}\left(\mathrm{tan}81\theta -\mathrm{tan}27\theta \right)$

$\therefore \frac{\mathrm{sin}\theta }{\mathrm{cos}3\theta }+\frac{\mathrm{sin}3\theta }{\mathrm{cos}9\theta }+\frac{\mathrm{sin}9\theta }{\mathrm{cos}27\theta }+\frac{\mathrm{sin}27\theta }{\mathrm{cos}81\theta }=\frac{1}{2}\left(\mathrm{tan}81\theta -\mathrm{tan}\theta \right)$

$⇒\frac{B-C}{A}=\frac{81+1}{1/2}=164$

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