How do you simplify

${\mathrm{tan}}^{2}4b-{\mathrm{sec}}^{2}4b$

Yahir Haas
2022-01-21
Answered

How do you simplify

${\mathrm{tan}}^{2}4b-{\mathrm{sec}}^{2}4b$

You can still ask an expert for help

bemolizisqt

Answered 2022-01-22
Author has **16** answers

Call $x=4b$

$f\left(x\right)=(\mathrm{tan}x-\mathrm{sec}x)(\mathrm{tan}x+\mathrm{sec}x)$

$=(\frac{se\in x}{\mathrm{cos}x}-\frac{1}{\mathrm{cos}x})(\frac{\mathrm{sin}x}{\mathrm{cos}x}+\frac{1}{\mathrm{cos}x})$

$=\frac{(\mathrm{sin}x-1)(\mathrm{sin}x+1)}{{\mathrm{cos}}^{2}x}=-\frac{1-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}$

$=-\frac{{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}=-1$

Jason Olsen

Answered 2022-01-23
Author has **14** answers

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$4b=x$

${\mathrm{tan}}^{2}\left(4b\right)+1={\mathrm{sec}}^{2}\left(4b\right)$

${\mathrm{tan}}^{2}\left(4b\right)-{\mathrm{sec}}^{2}\left(4b\right)=-1$

We know that

Let's let

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