Respuesta :
[tex]\bf \textit{Pythagorean Identities}
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sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
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[tex]\bf [sec(\theta )-tan(\theta )][1+sin(\theta )]\implies \left[ \frac{1}{cos(\theta )}-\frac{sin(\theta )}{cos(\theta )} \right][1+sin(\theta )] \\\\\\\ \left[\frac{1-sin(\theta )}{cos(\theta )} \right][1+sin(\theta )]\implies \cfrac{\stackrel{\textit{difference of squares}}{[1-sin(\theta )][1+sin(\theta )]}}{cos(\theta )} \\\\\\ \cfrac{1^2-sin^2(\theta )}{cos(\theta )}\implies \cfrac{cos^2(\theta )}{cos(\theta )}\implies cos(\theta )[/tex]
[tex]\bf [sec(\theta )-tan(\theta )][1+sin(\theta )]\implies \left[ \frac{1}{cos(\theta )}-\frac{sin(\theta )}{cos(\theta )} \right][1+sin(\theta )] \\\\\\\ \left[\frac{1-sin(\theta )}{cos(\theta )} \right][1+sin(\theta )]\implies \cfrac{\stackrel{\textit{difference of squares}}{[1-sin(\theta )][1+sin(\theta )]}}{cos(\theta )} \\\\\\ \cfrac{1^2-sin^2(\theta )}{cos(\theta )}\implies \cfrac{cos^2(\theta )}{cos(\theta )}\implies cos(\theta )[/tex]
