Q. The value of $\sin{20{^\circ}}(4 + \sec{20{^\circ}})$ is

A) 0
B) 1
C) $\sqrt{2}$
D) $\sqrt{3}$
Explanation

(d)
$\sin{20{^\circ}\left( 4 + \frac{1}{\cos{20{^\circ}}} \right) = \frac{4\sin{20{^\circ}\cos{20{^\circ} + \sin{20{^\circ}}}}}{\cos{20{^\circ}}}}$
$= \frac{\sin{40{^\circ}{+ sin}{40{^\circ} + \sin{20{^\circ}}}}}{\cos{20{^\circ}}}$
$= \frac{\sin{40{^\circ}{+ 2sin}{30{^\circ}\cos{10{^\circ}}}}}{\cos{20{^\circ}}}$
$= \frac{\cos{50{^\circ} + \cos{10{^\circ}}}}{\cos{20{^\circ}}}$
$\lbrack\because\sin{40{^\circ} = \cos{(90{^\circ} - 40{^\circ})}}\rbrack$
$= \frac{2\cos{30{^\circ}\cos{20{^\circ}}}}{\cos{20{^\circ}}} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.