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$\tan {20^ \circ } + \tan {40^ \circ } + \sqrt 3 {\text{ tan 2}}{{\text{0}}^ \circ }\tan {40^ \circ }$

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Hint- In this question, it is stated that we must find the value of this expression, so in order to solve this question we have to use the simple formula of $\tan \left( {A + B} \right)$ . This formula will help you to do the simplification of this expression given above -

__Complete step-by-step solution -__

In this question we have to find out the value of $\tan {20^ \circ }{\text{ + }}\tan {40^ \circ } + {\text{ }}\sqrt 3 \tan {20^ \circ }\tan {40^ \circ }$.

Now we can write $\left( {\tan {\text{ }}{{60}^ \circ }} \right)$ in the following manner -

$\tan {60^ \circ }$ = $\tan \left( {{{40}^ \circ } + {{20}^ \circ }} \right)$ ------(1)

As we know that:-

$\tan \left( {A + B} \right){\text{ = }}\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ -------(2)

Now putting (2) into (1) we will get,

$\sqrt 3 {\text{ = }}{\dfrac{{\tan {{40}^ \circ } + tan{{20}^ \circ }}}{{1 - {\text{ tan4}}{{\text{0}}^ \circ }\tan {{20}^ \circ }}}^{}}$ $\left( {\tan {{60}^ \circ } = {\text{ }}\sqrt 3 } \right)$

By cross multiplication, we will get,

$\sqrt 3 - \sqrt 3 {\text{ tan4}}{{\text{0}}^ \circ }{\text{ tan2}}{{\text{0}}^ \circ }{\text{ = tan4}}{{\text{0}}^ \circ }{\text{ + tan2}}{{\text{0}}^ \circ }$

Or $\tan {40^ \circ } + \tan {20^ \circ }{\text{ + }}\sqrt 3 \tan {\text{4}}{{\text{0}}^ \circ }\tan {20^ \circ }{\text{ = }}\sqrt 3 $

Thus, the value of the given expression is $\sqrt 3 $ .

Note- Whenever we face such types of problems, the key concept is that we have to use the formula of trigonometric functions. Here in this question we have to apply the formula of $\tan \left( {A + B} \right)$ by applying the formula we will get our required equation after that we have to put the value of $\tan {60^ \circ }$that is $\sqrt 3 $ . Next step is to cross multiply the equation of $\tan \left( {A + B} \right)$ with $\sqrt 3 $ like we did in the question and we will get our final answer.

In this question we have to find out the value of $\tan {20^ \circ }{\text{ + }}\tan {40^ \circ } + {\text{ }}\sqrt 3 \tan {20^ \circ }\tan {40^ \circ }$.

Now we can write $\left( {\tan {\text{ }}{{60}^ \circ }} \right)$ in the following manner -

$\tan {60^ \circ }$ = $\tan \left( {{{40}^ \circ } + {{20}^ \circ }} \right)$ ------(1)

As we know that:-

$\tan \left( {A + B} \right){\text{ = }}\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ -------(2)

Now putting (2) into (1) we will get,

$\sqrt 3 {\text{ = }}{\dfrac{{\tan {{40}^ \circ } + tan{{20}^ \circ }}}{{1 - {\text{ tan4}}{{\text{0}}^ \circ }\tan {{20}^ \circ }}}^{}}$ $\left( {\tan {{60}^ \circ } = {\text{ }}\sqrt 3 } \right)$

By cross multiplication, we will get,

$\sqrt 3 - \sqrt 3 {\text{ tan4}}{{\text{0}}^ \circ }{\text{ tan2}}{{\text{0}}^ \circ }{\text{ = tan4}}{{\text{0}}^ \circ }{\text{ + tan2}}{{\text{0}}^ \circ }$

Or $\tan {40^ \circ } + \tan {20^ \circ }{\text{ + }}\sqrt 3 \tan {\text{4}}{{\text{0}}^ \circ }\tan {20^ \circ }{\text{ = }}\sqrt 3 $

Thus, the value of the given expression is $\sqrt 3 $ .

Note- Whenever we face such types of problems, the key concept is that we have to use the formula of trigonometric functions. Here in this question we have to apply the formula of $\tan \left( {A + B} \right)$ by applying the formula we will get our required equation after that we have to put the value of $\tan {60^ \circ }$that is $\sqrt 3 $ . Next step is to cross multiply the equation of $\tan \left( {A + B} \right)$ with $\sqrt 3 $ like we did in the question and we will get our final answer.