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I have no clue where to start with this, I have been asked to show the following:

Let $0 < \theta < \frac{\pi}{2}$ and $f(x)\neq0$ on $[0, \theta]$ where$f$ is continuous on $[0, \theta]$. Show that there exists a positive constant K such that

$|f(x)|\ge K\tan{x}$ on $[0, \theta]$

My first thoughts were to try and use the definition of uniform continuity but I don't think that's the right tool to use.

gymbvghjkgkjkhgfkl
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dahaka5
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    $|f|$ attains a minimum on $[0,\theta]$, say $m$. Since $f>0$, $m>0$, hence $\frac{|f(x)|}{\tan(x)}\geq \frac m{\tan(x)}$.The function $x\mapsto \frac m{\tan(x)}$ is easily bounded below by $\frac m{\tan(\theta)}$. – Gabriel Romon Mar 08 '17 at 13:45

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The idea is that $f$ is continuous and non-zero on $[0,\theta]$. This means that it is bounded below by a strictly positive constant, as it attains its minimum. Thus $f(x) \geq m >0$. On the other hand, since $\theta< \pi/2$ the function $\tan$ is bounded above on $[0,\theta]$. Thus there exists $M$ such that $\tan x \leq M$ on $[0,\theta]$.

Now you have a function bounded below by $m>0$. It suffices to multiply it by $M/m$ to make it greater than $M$, and thus greater than the tangent on the same interval.

Beni Bogosel
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