Respuesta :
[tex]\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
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\cfrac{cot^2(x)}{csc(x)-1}=\cfrac{1+sin(x)}{sin(x)}\impliedby \textit{let's do the left-hand-side}[/tex]
[tex]\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)} \\\\\\ \cfrac{cos^2(x)}{sin(x)}\cdot \cfrac{1}{1-sin(x)}\implies \cfrac{cos^2(x)}{sin(x)[1-sin(x)]}[/tex]
[tex]\bf \cfrac{1-sin^2(x)}{sin(x)[1-sin(x)]}\implies \cfrac{1^2-sin^2(x)}{sin(x)[1-sin(x)]} \\\\\\ \cfrac{\underline{[1-sin(x)]}~[1+sin(x)]}{sin(x)\underline{[1-sin(x)]}}\implies \cfrac{1+sin(x)}{sin(x)}[/tex]
[tex]\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)} \\\\\\ \cfrac{cos^2(x)}{sin(x)}\cdot \cfrac{1}{1-sin(x)}\implies \cfrac{cos^2(x)}{sin(x)[1-sin(x)]}[/tex]
[tex]\bf \cfrac{1-sin^2(x)}{sin(x)[1-sin(x)]}\implies \cfrac{1^2-sin^2(x)}{sin(x)[1-sin(x)]} \\\\\\ \cfrac{\underline{[1-sin(x)]}~[1+sin(x)]}{sin(x)\underline{[1-sin(x)]}}\implies \cfrac{1+sin(x)}{sin(x)}[/tex]
