Question

If a trigonometric equation is given as $\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}}$ then $\theta =$ .
(a) $\dfrac{n\pi }{3}-\dfrac{\pi }{12},n\in z$
(b) $\dfrac{n\pi }{3}-\dfrac{\pi }{18},n\in z$
(c) $\dfrac{n\pi }{3}+\dfrac{\pi }{18},n\in z$
(d) $\dfrac{n\pi }{3}+\dfrac{\pi }{12},n\in z$

Answer 1

Hint: For solving this question we will use the formulas like $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ and $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ for proving $\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan 3\theta $ . After that, we will use the result of the solution of equation $\tan x=\tan y$ to write the final answer.

Complete step-by-step solution -
Given:
It is given that, $\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}}$ and we have to find the suitable values of $\theta $ .
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}..................\left( 1 \right) \\
 & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}..................\left( 2 \right) \\
 & \tan {{120}^{0}}=-\sqrt{3}......................................\left( 3 \right) \\
 & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}..........................\left( 4 \right) \\
 & \tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }........................\left( 5 \right) \\
\end{align}$
Now, we will be using the above formulas for solving this question.
Now, first, we will find the value of $\tan \left( {{120}^{0}}-\theta \right)$ in terms of $\tan \theta $ with the help of formula in the equation (2). Then,
$\tan \left( {{120}^{0}}-\theta \right)=\dfrac{\tan {{120}^{0}}-\tan \theta }{1+\tan {{120}^{0}}\tan \theta }$
Now, put $\tan {{120}^{0}}=-\sqrt{3}$ in the above equation from the equation (3). Then,
$\begin{align}
  & \tan \left( {{120}^{0}}-\theta \right)=\dfrac{\tan {{120}^{0}}-\tan \theta }{1+\tan {{120}^{0}}\tan \theta } \\
 & \Rightarrow \tan \left( {{120}^{0}}-\theta \right)=\dfrac{-\sqrt{3}-\tan \theta }{1-\sqrt{3}\tan \theta } \\
 & \Rightarrow \tan \left( {{120}^{0}}-\theta \right)=-\dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( 1-\sqrt{3}\tan \theta \right)} \\
 & \Rightarrow \tan \left( {{120}^{0}}-\theta \right)=\dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( \sqrt{3}\tan \theta -1 \right)}......................\left( 6 \right) \\
\end{align}$
Now, first, we will find the value of $\tan \left( {{120}^{0}}+\theta \right)$ in terms of $\tan \theta $ with the help of formula in the equation (1). Then,
$\tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan {{120}^{0}}+\tan \theta }{1-\tan {{120}^{0}}\tan \theta }$
Now, put $\tan {{120}^{0}}=-\sqrt{3}$ in the above equation from the equation (3). Then,
$\begin{align}
  & \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan {{120}^{0}}+\tan \theta }{1-\tan {{120}^{0}}\tan \theta } \\
 & \Rightarrow \tan \left( {{120}^{0}}+\theta \right)=\dfrac{-\sqrt{3}+\tan \theta }{1-\left( -\sqrt{3} \right)\tan \theta } \\
 & \Rightarrow \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta }.................\left( 7 \right) \\
\end{align}$
Now, we will solve for $\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)$ by substituting $\tan \left( {{120}^{0}}-\theta \right)=\dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( \sqrt{3}\tan \theta -1 \right)}$ from equation (6) and $\tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta }$ from equation (7). Then,
$\begin{align}
  & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right) \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( \sqrt{3}\tan \theta -1 \right)}\times \dfrac{\left( \tan \theta -\sqrt{3} \right)}{\left( 1+\sqrt{3}\tan \theta \right)} \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( \tan \theta +\sqrt{3} \right)\times \left( \tan \theta -\sqrt{3} \right)}{\left( \sqrt{3}\tan \theta +1 \right)\times \left( \sqrt{3}\tan \theta -1 \right)} \\
\end{align}$
Now, use the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ form equation (4) to write $\left( \tan \theta +\sqrt{3} \right)\times \left( \tan \theta -\sqrt{3} \right)={{\tan }^{2}}\theta -3$ and $\left( \sqrt{3}\tan \theta +1 \right)\times \left( \sqrt{3}\tan \theta -1 \right)=3{{\tan }^{2}}\theta -1$ in the above equation. Then,
$\begin{align}
  & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( \tan \theta +\sqrt{3} \right)\times \left( \tan \theta -\sqrt{3} \right)}{\left( \sqrt{3}\tan \theta +1 \right)\times \left( \sqrt{3}\tan \theta -1 \right)} \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( {{\tan }^{2}}\theta -3 \right)}{\left( 3{{\tan }^{2}}\theta -1 \right)} \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\left( {{\tan }^{3}}\theta -3\tan \theta \right)}{\left( 3{{\tan }^{2}}\theta -1 \right)} \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=-\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{\left( 3{{\tan }^{2}}\theta -1 \right)} \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)} \\
\end{align}$
Now, from equation (5) we know that, $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$ . Then,
$\begin{align}
  & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)} \\
 & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan 3\theta \\
\end{align}$
Now, as it is given that $\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}}$ . Then,
$\begin{align}
  & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan 3\theta \\
 & \Rightarrow \tan 3\theta =\dfrac{1}{\sqrt{3}} \\
\end{align}$
Now, we know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ . Then,
$\begin{align}
  & \tan 3\theta =\dfrac{1}{\sqrt{3}} \\
 & \Rightarrow \tan 3\theta =\tan \dfrac{\pi }{6}..........................\left( 8 \right) \\
\end{align}$
Now, before we proceed we should know one important result which we will use here.
If $\tan x=\tan y$ , then the general solution for $x$ in terms of y can be written as,
$x=n\pi +y....................\left( 9 \right)$ , where $n$ is any integer.
From equation (8) we know that, $\tan 3\theta =\tan \dfrac{\pi }{6}$ . So, we can use the formula from the equation (9). Then,
$\begin{align}
  & \tan 3\theta =\tan \dfrac{\pi }{6} \\
 & \Rightarrow 3\theta =n\pi +\dfrac{\pi }{6},n\in z \\
 & \Rightarrow \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{18},n\in z \\
\end{align}$
Now, from the above result, we conclude that, if $\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}}$ , then
$\theta =\dfrac{n\pi }{3}+\dfrac{\pi }{18},n\in z$ .
Hence, (c) will be the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to get the correct answer. And we should remember the results like $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$ and solution of the equation $\tan x=\tan y$ correctly and apply them without any mistake while solving. Moreover, we should avoid making calculation mistakes and after writing the final answer for such questions we should select the correct option only.