The value of $\sin (20 °) [4 + \sec (20 °)]$ is

$0$

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$1$

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$\sqrt{2}$

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$\sqrt{3}$

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Solution

The correct option is D
$\sqrt{3}$

Explanation for the correct option:
The given trigonometric expression: $\sin (20 °) [4 + \sec (20 °)]$ .
$\sin (20 °) [4 + \sec (20 °)] = \sin (20 °) [4 + \frac{1}{\cos (20 °)}] \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \sin (20 °) [\frac{4 \cos (20 °) + 1}{\cos (20 °)}] \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{4 \sin (20 °) \cos (20 °) + \sin (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{2 [2 \sin (20 °) \cos (20 °)] + \sin (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{2 \sin (40 °) + \sin (20 °)}{\cos (20 °)} [∵ \sin (2 A) = 2 \sin (A) \cos (B)] \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{2 \sin (60 ° - 20 °) + \sin (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{2 [\sin (60 °) \cos (20 °) - \cos (60 °) \sin (20 °)] + \sin (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{2 [(\frac{\sqrt{3}}{2}) \cos (20 °) - (\frac{1}{2}) \sin (20 °)] + \sin (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{\sqrt{3} \cos (20 °) - \sin (20 °) + \sin (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \frac{\sqrt{3} \cos (20 °)}{\cos (20 °)} \Rightarrow \sin (20 °) [4 + \sec (20 °)] = \sqrt{3}$
Therefore, the value of $\sin (20 °) [4 + \sec (20 °)]$ is $\sqrt{3}$ .
Hence, the correct option is (D).

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