Question

If $\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3$ then the general solution is_____
A. $\theta = (6n + 1)\dfrac{\pi }{{18}}$ , $\forall n \in I$
B. $\theta = (6n + 1)\dfrac{\pi }{9}$ , $\forall n \in I$
C. $\theta = (3n + 1)\dfrac{\pi }{9}$ , $\forall n \in I$
D. $\theta = (3n + 1)\dfrac{\pi }{{18}}$ , $\forall n \in I$

Answer 1

Hint : The solution generalized by means of periodicity is known as the general solution or the expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. If we have $\tan x = \tan y$ then the general solution is given by $x = n\pi + y$ . Now we have the equation $\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3$ , we convert this in the form of $\tan x = \tan y$ then we can apply the general solution.

Complete step-by-step answer :
Given,
$\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3$
Rearranging the terms, we get:
$\Rightarrow \tan \theta + \tan 2\theta = \sqrt 3 - \sqrt 3 \tan \theta \tan 2\theta$
Taking $\sqrt 3$ as common on the right hand side,
$\Rightarrow \tan \theta + \tan 2\theta = \sqrt 3 (1 - \tan \theta \tan 2\theta )$
Divide by $(1 - \tan \theta \tan 2\theta )$ on the both sides, we get:
$\Rightarrow \dfrac{{\tan \theta + \tan 2\theta }}{{(1 - \tan \theta \tan 2\theta )}} = \sqrt 3$
We know the formula $\tan (X + Y) = \dfrac{{\tan X + \tan Y}}{{(1 - \tan X.\tan Y)}}$ , comparing with right hand side of above equation we get:
$\left[ {X = \theta ,Y = 2\theta } \right]$ .
$\Rightarrow \tan (\theta + 2\theta ) = \sqrt 3$
But we know the standard value $\sqrt 3 = \tan \dfrac{\pi }{3}$ . Substituting this in above, we get: $\Rightarrow \tan (\theta + 2\theta ) = \tan \dfrac{\pi }{3}$
Adding the angel, we get:
$\Rightarrow \tan (3\theta ) = \tan \dfrac{\pi }{3}$
We know if we have $\tan x = \tan y$ then the general solution is $x = n\pi + y$ , comparing this with the above equation, we apply the general solution.
Then above equation becomes,
$\Rightarrow 3\theta = n\pi + \dfrac{\pi }{3}$
Divide by 3 on the both sides, we get:
$\Rightarrow \theta = \dfrac{{n\pi }}{3} + \dfrac{\pi }{9}$ , $\forall n \in I$
We can stop here, but the obtained answer does not match with any of the given options.
So taking $\dfrac{\pi }{9}$ as common,
$\Rightarrow \theta = \dfrac{\pi }{9}\left( {3n + 1} \right)$ , $\forall n \in I$
So, the correct answer is “Option C”.

Note : There is a difference between general solution and principle solution. The solutions lying between zero degree and 360 degree are known as principle solutions. Remember the general solution of six trigonometric functions. As all the general solutions of sine and cosine and tangent are look alike, be careful while applying the general solution.