Trigonometry : If tan aθ = tan bθ is a trigonometric equations and θ are in A.P. Then the general solution of θ is

Question

My try :

I have to proceed from here but I can't. So please tell me the next steps or any other process.

Answer 1

Thus, $$(a-b)\theta=\pi k,$$ where $k\in\mathbb Z$.

Now, consider two cases:

1. $a-b=0$;

2. $a-b\neq0$.

Can you end it now?

Remember that $\cos a\theta\neq0$ and $\cos b\theta\neq0$

Answer 2

It seems you are missing the signs or the fraction operations. It must be: $$\tan a\theta=\tan b\theta \Rightarrow \frac{\sin a\theta}{\cos a\theta}=\frac{\sin b\theta}{\cos b\theta} \Rightarrow \frac{\sin a\theta}{\cos a\theta}-\frac{\sin b\theta}{\cos b\theta}=0 \Rightarrow \\ \frac{\sin a\theta\cos b\theta-\sin b\theta\cos a\theta}{\cos a\theta\cos b\theta}=0 \Rightarrow \frac{\sin (a\theta-b\theta)}{\cos a\theta\cos b\theta}=0 \Rightarrow \\ \sin (a\theta-b\theta)=0, \cos a\theta\ne 0, \cos b\theta\ne 0.$$ Can you finish?