If P=tan⁡3n+1θ−tan⁡θ and Q=∑r=0n sin⁡3rθcos⁡3r+1θ then

Q=∑r=0n sin⁡3rθcos⁡3rθcos⁡3rθcos⁡3r+1θ=12∑r=0n tan⁡3r+1θ−tan⁡3rθ=12P

If P=tan⁡3n+1θ−tan⁡θ and Q=∑r=0n sin⁡3rθcos⁡3r+1θ then

Q=∑r=0n sin⁡3rθcos⁡3rθcos⁡3rθcos⁡3r+1θ=12∑r=0n tan⁡3r+1θ−tan⁡3rθ=12P

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If $P = (\tan (3^{n + 1} θ) - \tan θ) and Q = \sum_{r = 0}^{n} \frac{\sin (3^{r} θ)}{\cos (3^{r + 1} θ)}$ then

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P=3Q

2P=Q

3P=Q

P=2Q

answer is A.

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Detailed Solution

$Q = \sum_{r = 0}^{n} \frac{\sin (3^{r} θ) \cos (3^{r} θ)}{\cos (3^{r} θ) \cos (3^{r + 1} θ)} = \frac{1}{2} \sum_{r = 0}^{n} \tan (3^{r + 1} θ) - \tan (3^{r} θ) = \frac{1}{2} P$

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If P=tan⁡3n+1θ−tan⁡θ and Q=∑r=0n sin⁡3rθcos⁡3r+1θ then

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If $P = (\tan (3^{n + 1} θ) - \tan θ) and Q = \sum_{r = 0}^{n} \frac{\sin (3^{r} θ)}{\cos (3^{r + 1} θ)}$ then