Question

Solve the equation \[\tan \theta +\tan 2\theta =\tan 3\theta \].

Answer 1

Hint: We will use \[tan3\theta =\tan \left( \theta +2\theta \right)\] in the above question and we will substitute the value of \[\tan \left( \theta +2\theta \right)\] by using the trigonometric identity as follows:

Complete step-by-step answer:

\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\].

We have been given the equation \[\tan \theta +\tan 2\theta =\tan 3\theta \].

We know that \[tan3\theta =\tan \left( \theta +2\theta \right)\].

We also know that \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\].

So by using the above mentioned trigonometric identity in \[\tan \left( \theta +2\theta \right)\], we get as follows:

\[tan3\theta =\tan \left( \theta +2\theta \right)=\dfrac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }\].

Now considering \[tan3\theta =\dfrac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }\],

on cross multiplication, we get as follows:

\[tan3\theta \left( 1-\tan \theta \tan 2\theta \right)=\tan \theta +\tan 2\theta \].

Now we will substitute the value of \[\left( \tan \theta +\tan 2\theta \right)\] in the given equation, we get as follows:

\[tan3\theta \left( 1-\tan \theta \tan 2\theta \right)=\tan 3\theta \].

Taking \[\tan 3\theta \] on the left hand side of the equation, we get as follows:

\[tan3\theta \left( 1-\tan \theta \tan 2\theta \right)-\tan 3\theta =0\].

On taking \[\tan 3\theta \] as common, we get as follows:

\[\begin{align}

  & tan3\theta \left( 1-\tan \theta \tan 2\theta -1 \right)=0 \\

 & -\tan 3\theta \tan \theta \tan 2\theta =0 \\

 & \tan \theta \tan 2\theta tan3\theta =0 \\

 & \Rightarrow \tan \theta =0,\tan 2\theta =0,\tan 3\theta =0 \\

\end{align}\]

As we know the general solution for \[\tan \theta =\tan \theta \] is given by, \[\theta =n\pi

+\alpha \], where ‘n’ is any positive integer.

\[\begin{align}

  & \Rightarrow \tan \theta =\tan \theta \\

 & \therefore \theta =n\pi \\

 & \Rightarrow \tan 2\theta =\tan 0\Rightarrow 2\theta =n\pi \\

 & \therefore \theta =\dfrac{n\pi }{2} \\

 & \Rightarrow \tan 3\theta =\tan 0\Rightarrow 3\theta =n\pi \\

 & \therefore \theta =\dfrac{n\pi }{3} \\

\end{align}\]

Now on substituting these values ‘\[\theta \]’ in the given equation we observed that for \[\theta =\dfrac{n\pi }{2}\], we get the equation as \[\infty =\infty \]. So we cannot include this in solution.

Therefore, the solution of the given equation is \[\theta =n\pi \] and \[\theta =\dfrac{n\pi }{3}\].


Note: Check the value of \[\theta \] by substituting them in the given equation as there is a chance that we get an extra solution which doesn’t satisfy the equation and thus we have to exclude it from the final solution of the equation. Also, be careful of the sign while doing calculation.