# Why is $\frac{\sin x + \tan x}{\cos x + \cot x}$ always positive (when it's defined)?

Well, I've done my best to prove this, but I couldn't. I tried to open the functions, simplify them, but I just couldn't get to it. Well, here it goes.

Let $$0\leq x\leq 2\pi$$, and $$y = \displaystyle\frac{\sin x + \tan x}{\cos x + \cot x}$$ Prove that $$\;y>0\;$$ if $$\;x\neq k\displaystyle\frac{\pi}{2}, k\in\mathbb{Z}$$.

Could anyone help? Oh, and please: no Calculus.

## 2 Answers

$$y = \displaystyle\frac{\sin x\cos x(\sin x + \tan x)}{\sin x\cos x(\cos x + \cot x)}=\frac{\sin^2x(\cos x+1)}{\cos^2x(\sin x+1)}$$

$$\sin^2x$$, $$\cos^2x$$, $$\cos x+1$$ and $$\sin x+1$$ are all positive.

• That's pretty clever! Thanks a lot! – Italo Marinho May 2 at 2:44

$$y = \frac{sinx+tanx}{cosx+cotx} = \frac{sinx+\frac{sinx}{cosx}}{cosx+\frac{cosx}{sinx}} = \frac{sin^2xcosx+sin^2x}{cos^2xsinx+cos^2x} = \frac{sin^2x(1+cosx)}{cos^2x(1+sinx)}$$ $$\;x\neq k\displaystyle\frac{\pi}{2}, k\in\mathbb{Z}\;$$ $$\Rightarrow\;sinx\neq 0, \;sinx\neq \pm1, \;cosx\neq 0, \;cosx\neq \pm1$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Rightarrow\;sin^2x\gt 0, \;cos^2x\gt 0, \;1+cosx\gt 0, \;1+sinx\gt 0$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Rightarrow\;y\gt 0$$