If $p = \tan (3^{n + 1} θ) - \tan θ a n d Q = \sum_{}^{} \frac{\sin (3^{r} θ)}{\cos (3^{r + 1} θ)}$ , then

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$2 P = Q$

$P = 2 Q$

$P = 3 Q$

$3 P = Q$

answer is A.

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Detailed Solution

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

$= \frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} + \dots + \frac{\sin (3^{n} θ)}{\cos (3^{n + 1} θ)}$

As, $\frac{\sin θ}{\cos 3 θ} = \frac{2 \sin θ \cdot \cos θ}{2 \cos 3 θ \cdot \cos θ} = \frac{1}{2} [\frac{\sin 2 θ}{\cos 3 θ \cdot \cos θ}]$

$= \frac{1}{2} [\frac{\sin (3 θ - θ)}{\cos 3 θ \cdot \cos θ}] = \frac{1}{2} [\tan 3 θ - \tan θ]$

$∴ Q = \frac{1}{2} [(\tan 3 θ - \tan θ) + (\tan 9 θ - \tan 3 θ) + \dots + (\tan (3^{n + 1} θ) - \tan (3^{n} θ))]$

$= \frac{1}{2} [\tan (3^{n + 1} θ) - \tan θ]$

$Q = \frac{P}{2} \Rightarrow P = 2 Q$ .

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