# I need to simplify this expression. I'm not even sure where to start. tan &#x2061;<!-- ⁡ --

I need to simplify this expression. I'm not even sure where to start.
$\mathrm{tan}\left(\mathrm{arccos}\left(\frac{x}{4}\right)\right)$
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Jovan Wong
Let $y=\mathrm{tan}\left({\mathrm{cos}}^{-1}\left(\frac{x}{4}\right)\right)\phantom{\rule{thickmathspace}{0ex}},$, Now Put ${\mathrm{cos}}^{-1}\left(\frac{x}{4}\right)=\varphi \phantom{\rule{thickmathspace}{0ex}},$
Then $\frac{x}{4}=\mathrm{cos}\varphi \phantom{\rule{thickmathspace}{0ex}},$, Then $\mathrm{sec}\varphi =\frac{4}{x},$, So using $\mathrm{tan}\varphi =±\sqrt{{\mathrm{sec}}^{2}\varphi -1}=±\frac{\sqrt{16-{x}^{2}}}{x}$
So We get $y=\mathrm{tan}\varphi =±\frac{\sqrt{16-{x}^{2}}}{x}$
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varitero5w
Let $\mathrm{arccos}\frac{x}{4}=\theta$
This implies
$\frac{x}{4}=\mathrm{cos}\theta$
Forming a right angled triangle with the above and using pythagoras' theorem, opposite side to angle $\theta$ will be: $±\sqrt{16-{x}^{2}}$
Therefore,
$\mathrm{tan}\theta =\mathrm{tan}\left(\mathrm{arccos}\frac{x}{4}\right)=\frac{\sqrt{16-{x}^{2}}}{x}$