cos⁡3θcos3⁡θ+sin⁡3θsin3⁡θ is equal to

cos⁡3θcos3⁡θ+sin⁡3θsin3⁡θ=4cos3⁡θ−3cos⁡θcos3⁡θ+3sin⁡θ−4sin3⁡θsin3⁡θ=4×3cos2⁡θ−sin2⁡θ4sin2⁡θcos2⁡θ=12cos⁡2θsin⁡2θ×1sin⁡2θ

cos⁡3θcos3⁡θ+sin⁡3θsin3⁡θ is equal to

cos⁡3θcos3⁡θ+sin⁡3θsin3⁡θ=4cos3⁡θ−3cos⁡θcos3⁡θ+3sin⁡θ−4sin3⁡θsin3⁡θ=4×3cos2⁡θ−sin2⁡θ4sin2⁡θcos2⁡θ=12cos⁡2θsin⁡2θ×1sin⁡2θ

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$\frac{\cos 3 θ}{\cos^{3} θ} + \frac{\sin 3 θ}{\sin^{3} θ}$ is equal to

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$3 \cos 2 θ cosec 2 θ$

$3 \cot 2 θ cosec 2 θ$

$12 \cot 2 θ cosec 2 θ$

$12 \tan 2 θ \sec 2 θ$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{\cos 3 θ}{\cos^{3} θ} + \frac{\sin 3 θ}{\sin^{3} θ} \\ = \frac{4 \cos^{3} θ - 3 \cos θ}{\cos^{3} θ} + \frac{3 \sin θ - 4 \sin^{3} θ}{\sin^{3} θ} \\ = 4 \times 3 \frac{(\cos^{2} θ - \sin^{2} θ)}{4 \sin^{2} θ \cos^{2} θ} = \frac{12 \cos 2 θ}{\sin 2 θ} \times \frac{1}{\sin 2 θ} \end{array}$