Question

The value of $\tan {40^ \circ } + \tan {20^ \circ } + \sqrt 3 \tan {20^ \circ }\tan {40^ \circ }$ is
A.$\sqrt {12}$
B.$\dfrac{1}{{\sqrt 3 }}$
C.$\sqrt 3$
D.$\dfrac{{\sqrt 3 }}{2}$

Answer 1

Firstly, we will try to make the convert the problem to the formula
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Where A and B are the angles. Here, in this question, we are given two measures of the angles i.e. ${20^ \circ }$ and ${40^ \circ }$ . So, we will let one of the angles to be A and the other one to be B. Also, we know the value of $\tan \theta$ for $\theta = {0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$ . We will then solve the given problem by solving the obtained equation by using the above formula for $\tan (A + B)$ .

Complete answer:
We know,
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Let $A = {40^ \circ }$ and $B = {20^ \circ }$ in the above formula, we get
$\tan ({40^ \circ } + {20^ \circ }) = \dfrac{{\tan {{40}^ \circ } + \tan {{20}^ \circ }}}{{1 - \tan {{40}^ \circ }\tan {{20}^ \circ }}}$
$\tan ({60^ \circ }) = \dfrac{{\tan {{40}^ \circ } + \tan {{20}^ \circ }}}{{1 - \tan {{40}^ \circ }\tan {{20}^ \circ }}}$
$\sqrt 3 = \dfrac{{\tan {{40}^ \circ } + \tan {{20}^ \circ }}}{{1 - \tan {{40}^ \circ }\tan {{20}^ \circ }}}$ ($\tan {60^ \circ } = \sqrt 3$ )
$\dfrac{{\sqrt 3 }}{1} = \dfrac{{\tan {{40}^ \circ } + \tan {{20}^ \circ }}}{{1 - \tan {{40}^ \circ }\tan {{20}^ \circ }}}$
$\sqrt 3 (1 - \tan {40^ \circ }\tan {20^ \circ }) = \tan {40^ \circ } + \tan {20^ \circ }$ (Cross multiplying)
$\sqrt 3 - \sqrt 3 \tan {40^ \circ }\tan {20^ \circ } = \tan {40^ \circ } + \tan {20^ \circ }$ (Opening the brackets)
$\sqrt 3 = \tan {40^ \circ } + \tan {20^ \circ } + \sqrt 3 \tan {40^ \circ }\tan {20^ \circ }$
(Shifting one term to the right hand side)
Hence, we get
$\tan {40^ \circ } + \tan {20^ \circ } + \sqrt 3 \tan {40^ \circ }\tan {20^ \circ } = \sqrt 3$
Therefore, option (C) is the correct answer.

Note:
We need to be very careful while using the formulas. When we see two angles such that their sum or difference becomes one of the angles whose values are known then we will always first try to use the formula $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ . Also, while we are shifting the terms from one side to another, we need to take care of the signs especially, negative signs. When we first see the question, we need to interpret what trigonometric formula is to be used.