# How do you simplify \tan^2 4b-\sec^2 4b

How do you simplify
${\mathrm{tan}}^{2}4b-{\mathrm{sec}}^{2}4b$
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bemolizisqt
Call $x=4b$
$f\left(x\right)=\left(\mathrm{tan}x-\mathrm{sec}x\right)\left(\mathrm{tan}x+\mathrm{sec}x\right)$
$=\left(\frac{se\in x}{\mathrm{cos}x}-\frac{1}{\mathrm{cos}x}\right)\left(\frac{\mathrm{sin}x}{\mathrm{cos}x}+\frac{1}{\mathrm{cos}x}\right)$
$=\frac{\left(\mathrm{sin}x-1\right)\left(\mathrm{sin}x+1\right)}{{\mathrm{cos}}^{2}x}=-\frac{1-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}$
$=-\frac{{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}=-1$
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Jason Olsen
An alternate solution using the Pythagorean Identity:
We know that ${\mathrm{tan}}^{2}\left(x\right)+1={\mathrm{sec}}^{2}\left(x\right)$
Let's let $4b=x$
${\mathrm{tan}}^{2}\left(4b\right)+1={\mathrm{sec}}^{2}\left(4b\right)$
${\mathrm{tan}}^{2}\left(4b\right)-{\mathrm{sec}}^{2}\left(4b\right)=-1$
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