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# If $\tan \theta + \tan \left( {{{60}^ \circ } + \theta } \right) + \tan \left( {{{120}^ \circ } + \theta } \right) = 3$, then $\theta =$A) $\dfrac{{n\pi }}{3} + \dfrac{\pi }{{12}}$B) $n\pi + \dfrac{\pi }{4}$C) $2n\pi$D) $n\pi$

Last updated date: 15th Jun 2024
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Hint: We will first use the formula for tan (x + y) and then put in the general known values of tangent of some common angles. After that, we will just perform normal calculations and comparison and thus get the answer.

Complete step-by-step solution:
Let us first note down the formula which we will require in the solution of this question.
We have the formula as follows:-
$\Rightarrow \tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}$
Therefore taking the parts of questions: $\tan \left( {{{60}^ \circ } + \theta } \right)$ and $\tan \left( {{{120}^ \circ } + \theta } \right)$. Then, we will get:-
$\Rightarrow \tan \left( {{{60}^ \circ } + \theta } \right) = \dfrac{{\tan {{60}^ \circ } + \tan \theta }}{{1 - \tan {{60}^ \circ }.\tan \theta }}$
We already know that $\tan {60^ \circ } = \sqrt 3$. So, on putting this in above expression, we will get:-
$\Rightarrow \tan \left( {{{60}^ \circ } + \theta } \right) = \dfrac{{\sqrt 3 + \tan \theta }}{{1 - \sqrt 3 \tan \theta }}$ ………..……..(1)
Similarly, we will have:-
$\Rightarrow \tan \left( {{{120}^ \circ } + \theta } \right) = \dfrac{{\tan {{120}^ \circ } + \tan \theta }}{{1 - \tan {{120}^ \circ }.\tan \theta }}$
We already know that $\tan {120^ \circ } = \tan \left( {{{90}^ \circ } + {{30}^ \circ }} \right) = - \cot {30^ \circ } = - \sqrt 3$.
So, on putting this in above expression, we will get:-
$\Rightarrow \tan \left( {{{120}^ \circ } + \theta } \right) = \dfrac{{ - \sqrt 3 + \tan \theta }}{{1 + \sqrt 3 \tan \theta }}$ ………………….(2)
Now, we will put in the equations (1) and (2) in the LHS of the given equation which is $\tan \theta + \tan \left( {{{60}^ \circ } + \theta } \right) + \tan \left( {{{120}^ \circ } + \theta } \right) = 3$. We will then obtain the following expression:-
LHS: $\tan \theta + \tan \left( {{{60}^ \circ } + \theta } \right) + \tan \left( {{{120}^ \circ } + \theta } \right)$
$\Rightarrow \tan \theta + \dfrac{{\sqrt 3 + \tan \theta }}{{1 - \sqrt 3 \tan \theta }} + \dfrac{{ - \sqrt 3 + \tan \theta }}{{1 + \sqrt 3 \tan \theta }}$ (Using the equations (1) and (2))
Taking LCM to simplify it a bit, we will get:-
$\Rightarrow \tan \theta + \dfrac{{\left( {\sqrt 3 + \tan \theta } \right)\left( {1 + \sqrt 3 \tan \theta } \right) + \left( { - \sqrt 3 + \tan \theta } \right)\left( {1 - \sqrt 3 \tan \theta } \right)}}{{\left( {1 - \sqrt 3 \tan \theta } \right)\left( {1 + \sqrt 3 \tan \theta } \right)}}$
Simplifying the numerator to get the following expression:-
$\Rightarrow \tan \theta + \dfrac{{\sqrt 3 + 3\tan \theta + \tan \theta + \sqrt 3 {{\tan }^2}\theta - \sqrt 3 + 3\tan \theta + \tan \theta - \sqrt 3 {{\tan }^2}\theta }}{{\left( {1 - \sqrt 3 \tan \theta } \right)\left( {1 + \sqrt 3 \tan \theta } \right)}}$
Simplifying by combining the like terms in the numerator to get:-
$\Rightarrow \tan \theta + \dfrac{{8\tan \theta }}{{\left( {1 - \sqrt 3 \tan \theta } \right)\left( {1 + \sqrt 3 \tan \theta } \right)}}$
Now, we know that we have an identity given by $(a + b)(a - b) = {a^2} - {b^2}$. Using this in the above expression, we will then obtain the following equation:-
$\Rightarrow \tan \theta + \dfrac{{8\tan \theta }}{{1 - 3{{\tan }^2}\theta }}$
Now taking the LCM to combine these, we will get:-
$\Rightarrow \dfrac{{\tan \theta \left( {1 - 3{{\tan }^2}\theta } \right) + 8\tan \theta }}{{1 - 3{{\tan }^2}\theta }}$
Simplifying the numerator to get:-
$\Rightarrow \dfrac{{\tan \theta - 3{{\tan }^3}\theta + 8\tan \theta }}{{1 - 3{{\tan }^2}\theta }}$
Simplifying the numerator further to get:-
$\Rightarrow \dfrac{{9\tan \theta - 3{{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$
We can write this as:-
$\Rightarrow 3\left( {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$
Now, we know that: $\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$
So, we get the LHS as:-
$\therefore LHS = 3\tan 3\theta$
Now, we will put this equal to RHS which is given to be 3.
$\therefore 3\tan 3\theta = 3$
$\Rightarrow \tan 3\theta = 1$
$\Rightarrow \theta = \dfrac{{n\pi }}{3} + \dfrac{\pi }{{12}}$

$\therefore$ The correct option is (A).

Note: The students must commit to memory the following formulas:
$\Rightarrow \tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}$
$\Rightarrow \tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$
The students must also know that if they just take the answer as $\dfrac{\pi }{{12}}$ that will be the principal solution. Here, it is not mentioned that we need to find a principal solution, therefore, looking at the options we went for general solutions.