Answer

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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.

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