Question

Solve the following expression and obtain the value of $ \theta $
 $ \tan \left( \theta \right)+\tan \left( 2\theta \right)+\sqrt{3}\tan \left( \theta \right)\tan \left( 2\theta \right)=\sqrt{3} $
a) $ \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{3},\text{ }n\in Z $
b) $ \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{6},\text{ }n\in Z $
c) \[\theta =\dfrac{n\pi }{3}+\dfrac{\pi }{12},\text{ }n\in Z\]
d) $ \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{9},\text{ }n\in Z $

Answer 1

Hint: We see that in the given expression $ \tan \left( \theta \right) $ and $ \tan \left( 2\theta \right) $ are present. Therefore, we can use the formula for tangent of sum of angles and try to convert the given equation into that form to get the value of $ \tan \left( 3\theta \right) $ from which we can get the value of $ \theta $ .

Complete step-by-step answer:
The given expression is
 $ \begin{align}
  & \tan \left( \theta \right)+\tan \left( 2\theta \right)+\sqrt{3}\tan \left( \theta \right)\tan \left( 2\theta \right)=\sqrt{3} \\
 & \Rightarrow \tan \left( \theta \right)+\tan \left( 2\theta \right)=\sqrt{3}\left( 1-\tan \left( \theta \right)\tan \left( 2\theta \right) \right) \\
 & \Rightarrow \dfrac{\tan \left( \theta \right)+\tan \left( 2\theta \right)}{1-\tan \left( \theta \right)\tan \left( 2\theta \right)}=\sqrt{3}......(1.1) \\
\end{align} $
We know that the formula for $ \tan \left( a+b \right) $ is given by
 $ \tan \left( a+b \right)=\dfrac{\tan \left( a \right)+\tan \left( b \right)}{1-\tan \left( a \right)\tan \left( b \right)}.............(1.2) $
Therefore, using $ a=\theta $ and $ b=2\theta $ in (1.2), we obtain
 $ \tan \left( 3\theta \right)=\tan \left( 2\theta +\theta \right)=\dfrac{\tan \left( \theta \right)+\tan \left( 2\theta \right)}{1-\tan \left( \theta \right)\tan \left( 2\theta \right)} $
Using this expression in equation (1.1), we can rewrite it as
 $ \tan \left( 3\theta \right)=\sqrt{3}...............(1.3) $
However, we know that $ \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3} $ and as the tan function has a periodicity of $ \pi $ , we can write
 $ \begin{align}
  & \tan \left( 3\theta \right)=\tan \left( \dfrac{\pi }{3} \right) \\
 & \Rightarrow 3\theta =\dfrac{\pi }{3}+n\pi \\
 & \Rightarrow \theta =\dfrac{\pi }{9}+\dfrac{n\pi }{3} \\
\end{align} $
Which matches option (d) given in the question. Therefore, option(d) is the correct answer to this question.

Note: In this question, we could also have found out the value of $ \tan \left( \theta \right) $ by using the formula for $ \tan \left( 2\theta \right)=\dfrac{2\tan \left( \theta \right)}{1-{{\tan }^{2}}\left( \theta \right)} $ to replace $ \tan \left( 2\theta \right) $ by $ \tan \left( \theta \right) $ . We would then get a cubic equation, which we can solve for $ \tan \left( \theta \right) $ and obtain the value of $ \theta $ from it. However, this method would be very lengthy and difficult to solve and will give the same answer as obtained in the solution. Therefore, in these types of questions, the method used in the solution is a faster and easier approach to get the value of $ \theta $ .