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Q.

$\frac{\cos^{3} θ - \cos 3 θ}{\cos θ} + \frac{\sin^{3} θ + \sin 3 θ}{\sin θ} =$

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a

3

b

6

c

9

d

4

answer is A.

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Detailed Solution

\frac{\cos^{3} θ - \cos 3 θ}{\cos θ} + \frac{\sin^{3} θ + \sin 3 θ}{\sin θ}

= \frac{\cos^{3} θ - 4 \cos^{3} θ + 3 \cos θ}{\cos θ} + \frac{\sin^{3} θ + 3 \sin θ - 4 \sin^{3} θ}{\sin θ}

$= \frac{3 \cos θ - 3 \cos^{3} θ}{\cos θ} + \frac{3 \sin θ - 3 \sin^{3} θ}{\sin θ}$

$= 3 - 3 \cos^{2} θ + 3 - 3 \sin^{2} θ$

$= 6 - 3 (\cos^{2} θ + \sin^{2} θ)$

$= 6 - 3 (1)$

$= 3$

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