Question

# The most general value of $\theta$ which satisfies both the equations $\tan \theta = \sqrt 3$ and $\cos ec\theta = - \dfrac{2}{{\sqrt 3 }}$ is A) $n\pi + \dfrac{{4\pi }}{3}:n \in I$B) $n\pi + \dfrac{{2\pi }}{3}:n \in I$C) $2n\pi + \dfrac{{4\pi }}{3}:n \in I$D) $2n\pi + \dfrac{{2\pi }}{3}:n \in I$

Hint: This is the basic question of trigonometry so we have to find in which coordinate these values exist. And we use a simple conversion of $\cos ec\theta$ into terms of $\sin \theta$ . This conversion is we do just for our benefit purpose. And have to find the general solution for each equation and after that we take the intersection of them. And write the result in the form of a general equation by defining $n$.

In this we have to remember the basic coordinate system of all the trigonometric function and their basic general solution formula
Here we choose first equation $\tan \theta = \sqrt 3$
And solution for $\theta = \dfrac{\pi }{3},\dfrac{{4\pi }}{3},\dfrac{{7\pi }}{3}.....$
Now we take second equation $\cos ec\theta = - \dfrac{2}{{\sqrt 3 }}$
So we convert this in term of $\sin \theta$
By using formula $\cos ec\theta = \dfrac{1}{{\sin \theta }}$
So we have $\dfrac{1}{{\sin \theta }} = - \dfrac{2}{{\sqrt 3 }}$
And $\sin \theta = - \dfrac{{\sqrt 3 }}{2}$
Now we know that $\sin \theta$ is negative in $IIIrd$ and $ivth$ coordinate
So $\theta = \pi + \dfrac{\pi }{3},2\pi - \dfrac{\pi }{3}$ this value is repeated after every $n\pi$ interval.
$\theta = \dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$
Now the common value in both the equation is $\dfrac{{4\pi }}{3}$
So the general solution is $n\pi + \dfrac{{4\pi }}{3}:n \in I$
Option A is the correct answer.

Note: We have to choose only that value of $\theta$ which is common because this value is repeated after every $n\pi$ interval. And we can find this by using graphical methods .
For graphical method have follow this steps
* Make graph of each function
* Denote all the values and compare where both take the same value.