Alternative ways to solve $\frac{\sin{\theta}+1}{\cos{\theta}}\leq 1$? The inequality is:
$$ \frac{\sin{\theta}+1}{\cos{\theta}}\leq 1\quad \text{  with  } \cos{\theta}\neq0 \land 0\leq \theta\lt 2\pi$$
I've tried splitting it up into cases of $\theta$ that make $\cos{\theta}$ positive so: $0\leq\theta<\frac{\pi}{2}$ and $\frac{3\pi}{2} \lt\theta\lt 2\pi$ thus allowing the inequality to simply to:
$$ \sin{\theta}-\cos{\theta}\leq -1  $$
which after using the fact that $\sin{\theta}-\cos{\theta}=\sqrt{2}\sin{(\theta-\frac{\pi}{4})}$ and thus solving this inequality$$\sqrt{2}\sin{(\theta-\frac{\pi}{4})}\leq -1  \text{ for } 0\leq\theta<\frac{\pi}{2} \text{ or }\frac{3\pi}{2} \lt\theta\lt 2\pi $$
and this inequality
$$ \sqrt{2}\sin{(\theta-\frac{\pi}{4})}\geq -1  \text{ for } \frac{\pi}{2}\lt\theta<\frac{3\pi}{2} $$
which give the solutions:
$$ \theta=0 \text{ and } \frac{\pi }{2}<\theta<\frac{3 \pi }{2}\text{ and } \frac{3 \pi }{2}<\theta<2 \pi $$

I'm just curious if there's another way to do it that isn't quite so case based since it seems that I'm just solving it by plugging in values and testing. So I'm wondering if there's a better more insightful or efficient way to to this?


\begin{align}
1 + \sin \theta &= ( \cos (\theta /2) + \sin (\theta /2))^2 \\
\cos \theta &= ( \cos ^2 (\theta /2) - \sin ^2 (\theta /2) \\
\frac{1 + \sin \theta }{\cos \theta } &= \frac{ \cos (\theta /2) + \sin (\theta /2) }{ \cos (\theta /2) - \sin (\theta /2)} = \frac{ 1+ \tan (\theta /2)}{1 - \tan (\theta /2)} 
\end{align}
Therefore the given inequation reduces to $\frac{2 \tan (\theta /2)}{1 - \tan (\theta /2)} \leq 0$ whose solution is $\tan (\theta /2) \leq 0$ or $\tan (\theta /2) \geq 1$. 
