Question

Find the value of $\tan {{20}^{\circ }}+4\sin {{20}^{\circ }}$
(a) $\sqrt{3}$
(b) $\dfrac{1}{\sqrt{3}}$
(c) 1
(d) -1

Answer 1

Hint: Assume the given expression as E. First of all use the conversion $\tan x=\dfrac{\sin x}{\cos x}$ to simplify. Now, take LCM and use the trigonometric identity $2\sin a\cos a=\sin 2a$. Further, break the terms of the numerator and use the identity $\sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$ to simplify. Convert the sine function into the cosine function using the complementary angle formula $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ in the numerator and use the identity $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$. Cancel the common factors and use the values $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ to get the answer.

Complete step by step answer:
Here we have been asked to find the value of the expression $\tan {{20}^{\circ }}+4\sin {{20}^{\circ }}$. Let us assume the given expression as E, so we have,
$\Rightarrow E=\tan {{20}^{\circ }}+4\sin {{20}^{\circ }}$
Using the conversion formula $\tan x=\dfrac{\sin x}{\cos x}$ and taking the LCM to simplify we get,
$\begin{align}
  & \Rightarrow E=\dfrac{\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}+4\sin {{20}^{\circ }} \\
 & \Rightarrow E=\dfrac{\sin {{20}^{\circ }}+4\sin {{20}^{\circ }}\cos {{20}^{\circ }}}{\cos {{20}^{\circ }}} \\
 & \Rightarrow E=\dfrac{\sin {{20}^{\circ }}+2\times \left( 2\sin {{20}^{\circ }}\cos {{20}^{\circ }} \right)}{\cos {{20}^{\circ }}} \\
\end{align}$
Using the trigonometric identity $2\sin a\cos a=\sin 2a$ we get,
$\begin{align}
  & \Rightarrow E=\dfrac{\sin {{20}^{\circ }}+2\sin {{40}^{\circ }}}{\cos {{20}^{\circ }}} \\
 & \Rightarrow E=\dfrac{\left( \sin {{20}^{\circ }}+\sin {{40}^{\circ }} \right)+\sin {{40}^{\circ }}}{\cos {{20}^{\circ }}} \\
\end{align}$
Using the identity $\sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$ we get,
$\begin{align}
  & \Rightarrow E=\dfrac{2\sin \left( \dfrac{{{40}^{\circ }}+{{20}^{\circ }}}{2} \right)\cos \left( \dfrac{{{40}^{\circ }}-{{20}^{\circ }}}{2} \right)+\sin {{40}^{\circ }}}{\cos {{20}^{\circ }}} \\
 & \Rightarrow E=\dfrac{2\sin \left( {{30}^{\circ }} \right)\cos \left( {{10}^{\circ }} \right)+\sin {{40}^{\circ }}}{\cos {{20}^{\circ }}} \\
\end{align}$
Substituting the value $\sin {{30}^{\circ }}=\dfrac{1}{2}$ we get,
$\begin{align}
  & \Rightarrow E=\dfrac{2\times \dfrac{1}{2}\times \cos {{10}^{\circ }}+\sin {{40}^{\circ }}}{\cos {{20}^{\circ }}} \\
 & \Rightarrow E=\dfrac{\cos {{10}^{\circ }}+\sin {{40}^{\circ }}}{\cos {{20}^{\circ }}} \\
\end{align}$
Now, converting the sine function into the cosine function by using the complementary angle formula $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ in the numerator we get,
$\Rightarrow E=\dfrac{\cos {{10}^{\circ }}+\cos {{50}^{\circ }}}{\cos {{20}^{\circ }}}$
Using the identity $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$ we get,
$\begin{align}
  & \Rightarrow E=\dfrac{2\cos \left( \dfrac{{{50}^{\circ }}+{{10}^{\circ }}}{2} \right)\cos \left( \dfrac{{{50}^{\circ }}-{{10}^{\circ }}}{2} \right)}{\cos {{20}^{\circ }}} \\
 & \Rightarrow E=\dfrac{2\cos {{30}^{\circ }}\cos {{20}^{\circ }}}{\cos {{20}^{\circ }}} \\
\end{align}$
Using the value $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and cancelling the common factor we get,
$\begin{align}
  & \Rightarrow E=2\times \dfrac{\sqrt{3}}{2} \\
 & \therefore E=\sqrt{3} \\
\end{align}$

So, the correct answer is “Option a”.

Note: Note that in the end you can also convert the cosine function into the sine function and then apply the suitable trigonometric identity to get the answer. But in that case you have to apply the complementary angle formula again to cancel the common factor in the denominator. You must remember all the trigonometric identities as they are used in other chapters and subjects also.