Question

Solve the given trigonometric equation:
$\tan \theta + \tan 2\theta + \tan 3\theta = 0.$

Answer 1

Hint: In order to solve the question, we will simply use the formula of $\tan \left( {A + B} \right)$ from the trigonometric table and we will make two factors by solving this factors we will get the value of $\theta $ and the following formula will be used to solve this question.
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}{\text{ and }}\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Complete step-by-step solution -
The given equation is
$\tan \theta + \tan 2\theta + \tan 3\theta = 0.$
Now simplifying the above equations using the formula
$
   \Rightarrow \tan \theta + \tan 2\theta + \tan 3\theta = 0 \\
   \Rightarrow \tan \theta + \tan 2\theta + \tan \left( {\theta + 2\theta } \right) = 0 \\
$
Using the formula $\left[ {\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right]$

\[
   \Rightarrow \tan \theta + \tan 2\theta + \dfrac{{\tan \theta + \tan 2\theta }}{{1 - \tan \theta \tan 2\theta }} = 0 \\
   \Rightarrow \left( {\tan \theta + \tan 2\theta } \right)\left( {1 - \tan \theta \tan 2\theta } \right) + \left( {\tan \theta + \tan 2\theta } \right) = 0 \\
   \Rightarrow \left( {\tan \theta + \tan 2\theta } \right)\left( {2 - \tan \theta \tan 2\theta } \right) =0 \\
   \Rightarrow \tan \theta + \tan 2\theta = 0{\text{ or }}\tan \theta \tan 2\theta = 2 \\
\]
If $\tan \theta + \tan 2\theta = 0$ then the general equation is give as
$
  \tan 2\theta = - \tan \theta = \tan \left( { - \theta } \right) \\
   \Rightarrow 2\theta = n\pi + \left( { - \theta } \right),n \in z \\
   \Rightarrow 3\theta = n\pi ,n \in z \\
   \Rightarrow \theta = \dfrac{{n\pi }}{3},n \in z \\
$
If \[\tan \theta \tan 2\theta = 2,\] then the general equation is given as
$\tan \theta \tan 2\theta = 2$
Using the formula $\left[ {\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}} \right]$
\[
   \Rightarrow \tan \theta \times \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = 2 \\
   \Rightarrow {\tan ^2}\theta = 1 - {\tan ^2}\theta \\
   \Rightarrow {\tan ^2}\theta = \dfrac{1}{2} \\
   \Rightarrow \tan \theta = \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow \tan \theta = \tan \left( {{{35}^0}16'} \right) \\
   \Rightarrow \theta = n\pi \pm {35^0}16',n \in z \\
\]
Therefore the solution set is given as
$ = \left\{ {\dfrac{{n\pi }}{3}:n \in z} \right\} \cup \left\{ {n\pi \pm {{35}^0}16':n \in z} \right\}$

Note: In order to solve these types of questions, learn all the formulas of trigonometric identities and learn how to use the values from trigonometric tables. Remember some important values. You must have the concept of quadrants and how to find the general solution of the trigonometric equations. Always try to solve the problems in steps.