Question

If $\cot \theta +\tan \theta =2\csc \theta$, then find the general value of $\theta$.

Answer 1

In the given question, we are given a trigonometric expression using which we have to find the general value of $\theta$. We will first write the trigonometric functions in terms of sine and cosine functions and reduce the expression as much as possible. Then, the obtained expression will look similar to $\sin \theta \left( 1-2\cos \theta \right)=0$. We will then separate out the components and find the general values of $\theta$ for these functions and later on, we will club them together. Hence, we will have the general value of $\theta$.

Complete step by step answer:
According to the given question, we are given an expression based on trigonometric functions. We are asked to find the general value of $\theta$.
The expression we have is,
$\cot \theta +\tan \theta =2\csc \theta$----(1)
In equation (1), we have the cotangent function, tangent function and the cosecant function, we will first write these functions in terms of sine and cosine functions. We will get,
$\Rightarrow \dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta }=\dfrac{2}{\sin \theta }$
We will now take the LCM in the LHS and we get,
$\Rightarrow \dfrac{\cos \theta .\cos \theta }{\sin \theta \cos \theta }+\dfrac{\sin \theta .\sin \theta }{\cos \theta \sin \theta }=\dfrac{2}{\sin \theta }$
Computing the functions, we get,
$\Rightarrow \dfrac{{{\cos }^{2}}\theta }{\sin \theta \cos \theta }+\dfrac{{{\sin }^{2}}\theta }{\cos \theta \sin \theta }=\dfrac{2}{\sin \theta }$
Adding up the numerators in the LHS, we get,
$\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\sin \theta \cos \theta }=\dfrac{2}{\sin \theta }$
We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$, so we have,
$\Rightarrow \dfrac{1}{\sin \theta \cos \theta }=\dfrac{2}{\sin \theta }$----(2)
We will now cross multiply in equation (2), we get,
$\Rightarrow \sin \theta =2\sin \theta \cos \theta$----(3)
Now bringing the functions to one side, we have,
$\Rightarrow \sin \theta -2\sin \theta \cos \theta =0$-----(4)
In equation (4), we can take the sine function common, so we get,
$\Rightarrow \sin \theta \left( 1-2\cos \theta \right)=0$----(5)
Separating out the components in the equation (5), we get,
$\sin \theta =0$ and $1-2\cos \theta =0$
Sine function gives a value 0 for all factors of $\pi$, so we can write,
$\theta =n\pi$, where n is an integer.
$1-2\cos \theta =0$
Solving this we get,
$\Rightarrow 2\cos \theta =1$
$\Rightarrow \cos \theta =\dfrac{1}{2}$
Cosine function gives a value $\dfrac{1}{2}$ for all $\dfrac{\pi }{3}$ in the first and the fourth quadrant. So, we have,
$\theta =2n\pi \pm \dfrac{\pi }{3}$

Therefore, the general value of $\theta =\left( n\pi \right)\cup \left( 2n\pi \pm \dfrac{\pi }{3} \right)$.

Note: We wrote the given expression in terms of sine and cosine function because it would have got very difficult if we would have tried solving the expression in tangent and cotangent form. Also, the values of sine and cosine functions should not be confused. Since, sine and cosine functions have a slight difference in the values at various angles so it should be taken care of.