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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.

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

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

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  • $\begingroup$ That's pretty clever! Thanks a lot! $\endgroup$ – Italo Marinho May 2 at 2:44
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$$y = \frac{sinx+tanx}{cosx+cotx} = \frac{sinx+\frac{sinx}{cosx}}{cosx+\frac{cosx}{sinx}} = \frac{sin^2xcosx+sin^2x}{cos^2xsinx+cos^2x} = \frac{sin^2x(1+cosx)}{cos^2x(1+sinx)}$$ $\;x\neq k\displaystyle\frac{\pi}{2}, k\in\mathbb{Z}\;$ $\Rightarrow\;sinx\neq 0, \;sinx\neq \pm1, \;cosx\neq 0, \;cosx\neq \pm1$

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

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

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