# Sin(theta)/1+cos(theta)=1-cos(theta)/sin(theta) can someone help?

May 10, 2018

$\theta \ne n \pi , n \in \mathbb{Z}$

#### Explanation:

First, please use math template, it is very frustrating to read something like this, and ask a precise question, I assume you want to find theta but I am not a mind reader.

$\frac{\sin \left(\theta\right)}{1 + \cos \left(\theta\right)} = \frac{1 - \cos \left(\theta\right)}{\sin \left(\theta\right)}$

We can suppose $1 + \cos \left(\theta\right)$ and $\sin \left(\theta\right)$ to be nonzero, otherwise the expression does not make sense, so we can multiply by $1 + \cos \left(\theta\right) \sin \left(\theta\right)$ and get

${\sin}^{2} \left(\theta\right) = \left(1 - \cos \left(\theta\right)\right) \left(1 + \cos \left(\theta\right)\right) = 1 - {\cos}^{2} \left(\theta\right) = {\sin}^{2} \left(\theta\right)$

So we get the tautology ${\sin}^{2} \left(\theta\right) = {\sin}^{2} \left(\theta\right)$, true for all $\theta$ that make sense, so we just need to exclude the ones such that $\sin \left(\theta\right) = 0$ and $\cos \left(\theta\right) = - 1$. So $\theta \ne n \pi$ and $\theta \ne \left(2 n + 1\right) \pi$, in particular since the second is weaker then the first we conclude.