Question

The value of $\sin {{20}^{\circ }}\left( 4+\sec {{20}^{\circ }} \right)$ is
$1)\text{ }0$
$2)\text{ 1}$
$3)\text{ }\sqrt{2}$
$4)\text{ }\sqrt{3}$

Answer 1

Hint: In this question we have been given with a trigonometric expression for which we have to find the value of the expression. We will solve this expression by writing the expression and simplifying it. We will use the half-angle formula which is given as $\sin 2\theta =2\sin \theta \cos \theta $. We will then use the expansion formula of $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ and then simplify the expression to get the required solution.

Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow \sin {{20}^{\circ }}\left( 4+\sec {{20}^{\circ }} \right)$
We know that $\cos x=\dfrac{1}{\sec x}$ therefore, on substituting, we get:
$\Rightarrow \sin {{20}^{\circ }}\left( 4+\dfrac{1}{\cos {{20}^{\circ }}} \right)$
On multiplying the terms, we get:
$\Rightarrow 4\sin {{20}^{\circ }}+\dfrac{\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}$
On taking the lowest common multiple, we get:
$\Rightarrow \dfrac{4\sin {{20}^{\circ }}\cos {{20}^{\circ }}+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}$
We can write the expression as:
$\Rightarrow \dfrac{2\left( 2\sin {{20}^{\circ }}\cos {{20}^{\circ }} \right)+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}$
On using the formula $\sin 2\theta =2\sin \theta \cos \theta $, we get:
$\Rightarrow \dfrac{2\sin {{40}^{\circ }}+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}$
Now we can expression ${{40}^{\circ }}$ as ${{60}^{\circ }}-{{40}^{\circ }}$ therefore, on substituting, we get:
\[\Rightarrow \dfrac{2\sin \left( {{60}^{\circ }}-{{40}^{\circ }} \right)+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}\]
Now on using the formula $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$, we get:
\[\Rightarrow \dfrac{2\sin {{60}^{\circ }}\cos {{20}^{\circ }}-2\cos {{60}^{\circ }}\sin {{20}^{\circ }}+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}\]
Now we know that $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\cos {{60}^{\circ }}=\dfrac{1}{2}$ therefore, on substituting, we get:
\[\Rightarrow \dfrac{2\left( \dfrac{\sqrt{3}}{2} \right)\cos {{20}^{\circ }}-2\left( \dfrac{1}{2} \right)\sin {{20}^{\circ }}+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}\]
On simplifying, we get:
\[\Rightarrow \dfrac{\sqrt{3}\cos {{20}^{\circ }}-\sin {{20}^{\circ }}+\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}\]
On subtracting the similar terms, we get:
\[\Rightarrow \dfrac{\sqrt{3}\cos {{20}^{\circ }}}{\cos {{20}^{\circ }}}\]
On simplifying, we get:
\[\Rightarrow \sqrt{3}\], which is the required value.
Therefore, we can write $\sin {{20}^{\circ }}\left( 4+\sec {{20}^{\circ }} \right)=\sqrt{3}$, which is the required solution.

Note: It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken. The various trigonometric identities and formulae should be remembered while doing these types of sums. The various Pythagorean identities should also be remembered while doing these types of questions.