The sum of the series, sin⁡θsec⁡3θ+sin⁡3θsec⁡32θ+sin⁡32θsec⁡33θ+…… upto n terms, is

sin⁡θ⋅sec⁡3θ+sin⁡3θ⋅sec⁡32θ+sin⁡32θ⋅sec⁡33θ+……. upto n terms =∑r=1n sin⁡3r−1θ⋅sec⁡3rθ=∑r=1n 2cos⁡3r−1θsin⁡3r−1θ2cos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n sin⁡2⋅3r−1θcos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n sin⁡3rθ−3r−1θcos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n sin⁡3rθ⋅cos⁡3r−1θ−cos⁡3rθ⋅sin⁡3r−1θcos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n tan⁡3rθ−tan⁡3r−1θ

The sum of the series, sin⁡θsec⁡3θ+sin⁡3θsec⁡32θ+sin⁡32θsec⁡33θ+…… upto n terms, is

sin⁡θ⋅sec⁡3θ+sin⁡3θ⋅sec⁡32θ+sin⁡32θ⋅sec⁡33θ+……. upto n terms =∑r=1n sin⁡3r−1θ⋅sec⁡3rθ=∑r=1n 2cos⁡3r−1θsin⁡3r−1θ2cos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n sin⁡2⋅3r−1θcos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n sin⁡3rθ−3r−1θcos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n sin⁡3rθ⋅cos⁡3r−1θ−cos⁡3rθ⋅sin⁡3r−1θcos⁡3r−1θ⋅cos⁡3rθ=12∑r=1n tan⁡3rθ−tan⁡3r−1θ

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The sum of the series, $\sin θsec 3 θ + \sin 3 θsec 3^{2} θ + \sin 3^{2} θsec 3^{3} θ + \dots \dots$ upto n terms, is

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$\frac{1}{2} [\tan 3^{n} θ - \tan 3^{n - 1} θ]$

$[\tan 3^{n} θ - \tan θ]$

$\frac{1}{2} [\tan 3^{n} θ - \tan θ]$

$\frac{1}{2} (\tan 3^{n} θ - 1)$

answer is C.

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Detailed Solution

$\begin{array}{l} \sin θ \cdot \sec 3 θ + \sin 3 θ \cdot \sec 3^{2} θ \\ + \sin 3^{2} θ \cdot \sec 3^{3} θ + \dots \dots . upto n terms \\ = \sum_{r = 1}^{n} \sin 3^{r - 1} θ \cdot \sec 3^{r} θ = \sum_{r = 1}^{n} \frac{2 \cos 3^{r - 1} θsin 3^{r - 1} θ}{2 \cos 3^{r - 1} θ \cdot \cos 3^{r} θ} \\ = \frac{1}{2} \sum_{r = 1}^{n} \frac{\sin (2 \cdot 3^{r - 1} θ)}{\cos 3^{r - 1} θ \cdot \cos 3^{r} θ} = \frac{1}{2} \sum_{r = 1}^{n} \frac{\sin (3^{r} θ - 3^{r - 1} θ)}{\cos 3^{r - 1} θ \cdot \cos 3^{r} θ} \\ = \frac{1}{2} \sum_{r = 1}^{n} \frac{\sin 3^{r} θ \cdot \cos 3^{r - 1} θ - \cos 3^{r} θ \cdot \sin 3^{r - 1} θ}{\cos 3^{r - 1} θ \cdot \cos 3^{r} θ} \\ = \frac{1}{2} \sum_{r = 1}^{n} (\tan 3^{r} θ - \tan 3^{r - 1} θ) \end{array}$