Question

Solve for the general value of theta $\theta $ :
$\tan \left( \theta \right)\tan \left( {{{120}^0} - \theta } \right)\tan \left( {{{120}^0} + \theta } \right) = \dfrac{1}{{\sqrt 3 }}$ is;

Answer 1

Hint: The above question deals with the simplification of trigonometric functions using standard identities along with some basic algebraic rules. To proceed with the question , the standard formulae of tangent function should be kept in mind . Notice here that the formula for tangent function can be used for simplification i.e. $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$ and $\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$.

Complete step by step answer:
The given question is;
$ \Rightarrow \tan \left( \theta \right)\tan \left( {{{120}^0} - \theta } \right)\tan \left( {{{120}^0} + \theta } \right) = \dfrac{1}{{\sqrt 3 }}$
Expanding $\tan \left( {{{120}^0} + \theta } \right)$ and $\tan \left( {{{120}^0} + \theta } \right)$ by the formulae defined above, we get;
$ \Rightarrow \tan \theta \left[ {\dfrac{{\tan \left( {{{120}^0}} \right) + \tan \theta }}{{1 - \tan \left( {{{120}^0}} \right)\tan \theta }}} \right]\left[ {\dfrac{{\tan \left( {{{120}^0}} \right) - \tan \theta }}{{1 + \tan \left( {{{120}^0}} \right)\tan \theta }}} \right] = \dfrac{1}{{\sqrt 3 }}$ $......\left( 1 \right)$
We know that , $\tan \left( {{{120}^0}} \right) = \tan \left( {{{180}^0} - {{60}^0}} \right)$
By the formula $\tan \left( {180 - \theta } \right) = - \tan \theta $
$\therefore \tan \left( {{{180}^0} - {{60}^0}} \right) = - \tan \left( {{{60}^0}} \right) = - \sqrt 3 $
$\left( {\because \tan {{60}^0} = \sqrt 3 } \right)$
Replacing the value of $\tan \left( {{{120}^0}} \right)$ in equation $\left( 1 \right)$, we get;
$ \Rightarrow \tan \theta \left[ {\dfrac{{ - \sqrt 3 + \tan \theta }}{{1 + \sqrt 3 \tan \theta }}} \right]\left[ {\dfrac{{ - \sqrt 3 - \tan \theta }}{{1 - \sqrt 3 \tan \theta }}} \right] = \dfrac{1}{{\sqrt 3 }}$
The above equation resembles with the formula of ${a^2} - {b^2} = \left( {a + b} \right)(a - b)$
Therefore, it can be reduced to;
$ \Rightarrow \tan \theta \left[ {\dfrac{{{{\left( { - \sqrt 3 } \right)}^2} - \left( {{{\tan }^2}\theta } \right)}}{{1 - {{\left( {\sqrt 3 \tan \theta } \right)}^2}}}} \right] = \dfrac{1}{{\sqrt 3 }}$
Simplifying the above equation;
$ \Rightarrow \tan \theta \left[ {\dfrac{{3 - {{\tan }^2}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right] = \dfrac{1}{{\sqrt 3 }}$
Multiplying $\tan \theta $ inside, we get;
$ \Rightarrow \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = \dfrac{1}{{\sqrt 3 }}$
By the formula, $\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^2}\theta }}{{1 - 3{{\tan }^2}\theta }}$
Applying the above formula;
$ \Rightarrow \tan \left( {3\theta } \right) = \dfrac{1}{{\sqrt 3 }}$
We know that, $\tan {60^0} = \tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}$
$ \Rightarrow \tan 3\theta = \tan \left( {\dfrac{\pi }{6}} \right)$ $......\left( 2 \right)$
According to the general solution condition of tangent function;
$ \Rightarrow \tan x = \tan \phi $
$ \Rightarrow x = n\pi + \phi $ , where $n \in z$ ( $z$ is set of integers)
(The general solution determines the value of $x$ which satisfies the equation)
Therefore equation $\left( 2 \right)$ reduces to;
$ \Rightarrow 3\theta = n\pi + \dfrac{\pi }{6}$ where $n \in z$
Simplifying further:
$ \Rightarrow \theta = \dfrac{{n\pi }}{3} + \dfrac{\pi }{{18}}$ where $n \in z$
Or $\theta $ can also be written as;
$ \Rightarrow \theta = \left( {6n + 1} \right)\dfrac{\pi }{{18}}$ where $n \in z$
Therefore the correct answer for this question is $\theta = \left( {6n + 1} \right)\dfrac{\pi }{{18}}$ where $n \in z$ .

Note: To solve this question, the concept of general solution of tangent function is used i.e. For
$\tan x = \tan \phi $ , the general solution is given by $x = n\pi + \phi $ . Although one condition is there i.e. $x$ and $\phi $ can not take the value as $\dfrac{\pi }{2}$ , because the value of tangent function for $\dfrac{\pi }{2}$ is not defined . There is also a concept of principal solution of a trigonometric identity which covers all values from $0{\text{ to 36}}{{\text{0}}^0}$ or
${0^0}{\text{ to 2}}\pi $ since all the trigonometric functions repeat themselves after a certain time
period. The principal solution involves the variables while the general solution involves the integer $n$ .