Question

If $ \tan \theta +\tan 2\theta +\sqrt{3}\tan \theta \tan 2\theta =\sqrt{3} $ , then \[\]
A. $ \theta =\left( 6n+1 \right)\dfrac{\pi }{18},\forall n\in I $ \[\]
B. $ \theta =\left( 6n+1 \right)\dfrac{\pi }{9},\forall n\in I $ \[\]
C. $ \theta =\left( 3n+1 \right)\dfrac{\pi }{9},\forall n\in I $ \[\]
D. $ \theta =\left( 6n+1 \right)\dfrac{\pi }{18},\forall n\in I $ \[\]

Answer 1

Hint: We take $ \sqrt{3}\tan \theta \tan 2\theta $ to the right hand side, take $ \sqrt{3} $ common and divide both sides by $ 1-\tan \theta \tan 2\theta $ . We express the obtained expression at the left hand side in the form of $ \tan \left( A+B \right) $ suing the formula $ \dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}=\tan \left( A+B \right) $ . We find the solutions of the equation in tangent $ \tan x=\tan \alpha $ as $ x=n\pi +\alpha $ .

Complete step-by-step answer:
We know from tangent sum of the angles formula that for any two angles with measures $ A $ and $ B $ we have
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\cdot \tan B}\]
We can also find the value of $ \tan {{60}^{\circ }} $ by converting it to the ratio of sine and cosine. We have
\[\tan {{60}^{\circ }}=\dfrac{\sin {{60}^{\circ }}}{\cos {{60}^{\circ }}}=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}=\sqrt{3}\]
We convert $ {{60}^{\circ }} $ into radian and have
\[\begin{align}
  & {{60}^{\circ }}=\dfrac{{{60}^{\circ }}}{{{180}^{\circ }}}\times \pi =\dfrac{\pi }{3} \\
 & \Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\tan \left( {{60}^{\circ }} \right)=\sqrt{3} \\
\end{align}\]
We also know that the solutions of the equation $ \tan x=\tan \alpha $ (where $ x $ is the unknown variable and $ \alpha $ is measure of angle) are given by
\[x=n\pi +\alpha \]
Here $ n $ is any integer , so we have $ n\in I $ .

We are given the question a trigonometric equation in tangent of any acute angle $ \theta $ as
\[\tan \theta +\tan 2\theta +\sqrt{3}\tan \theta \tan 2\theta =\sqrt{3}\]
Let us collect the terms having $ \sqrt{3} $ at the right hand side of the equation. We have
\[\Rightarrow \tan \theta +\tan 2\theta =\sqrt{3}-\sqrt{3}\tan \theta \tan 2\theta \]
We take $ \sqrt{3} $ common in the right hand side of the equation and have
\[\Rightarrow \tan \theta +\tan 2\theta =\sqrt{3}\left( 1-\tan \theta \tan 2\theta \right)\]
Let us divide both side of the above equation by \[1-\tan \theta \tan 2\theta \] and have,
\[\Rightarrow \dfrac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }=\sqrt{3}\]
We observe that the expression in tangent of angles in the left hand side is in the form tangent sum of two angles $ \tan \left( A+B \right) $ where $ A=\theta ,B=2\theta $ . We use the formula and proceed to have
\[\begin{align}
  & \Rightarrow \tan \left( \theta +2\theta \right)=\sqrt{3} \\
 & \Rightarrow \tan \left( 3\theta \right)=\sqrt{3} \\
\end{align}\]
We know that $ \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3} $ . We put $ \tan \left( \dfrac{\pi }{3} \right) $ in place of $ \sqrt{3} $ and have,
\[\Rightarrow \tan \left( 3\theta \right)=\tan \left( \dfrac{\pi }{3} \right)\]
We can obtain the solution of the above equation by taking $ x=3\theta ,\alpha =\dfrac{\pi }{3} $ and for some $ n\in I $ . We have,
\[\Rightarrow 3\theta =n\pi +\dfrac{\pi }{3}\]
We divide both side of the equation by 3 and have
\[\Rightarrow \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{9}\]
We take $ \dfrac{\pi }{9} $ common in the right hand side of the equation and have,
\[\Rightarrow \theta =\dfrac{\pi }{9}\left( 3n+1 \right),n\in I\]

So, the correct answer is “Option C”.

Note: We note that $ 1-\tan 2\theta \tan \theta $ cannot be zero here and that is why we could divide with it. The value of $ \theta $ for which $ 1-\tan 2\theta \tan \theta =0 $ will not satisfy the given equation. Here we can find $ \theta =\left( \dfrac{2n+1}{6} \right)\pi $ .If $ A=B $ we get tangent double angle formula $ \tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A} $ . The tangent difference angle formula is given by $ \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\cdot \tan B} $