Question

Mostre que $\frac{ 1-2cos^{2}x+ cos^{4}x }{1-2 sen^{2}x+ sen^{4}x } = tg^{4}x$ (Sugestão: O numerador e o denominador são trinômio quadrado perfeito, fatore!)

Answer 1

$\frac{1-2xos^2x+cos^4x}{1-2sen^2x+sen^4x}=tg^4x\\ \\ \\ \frac{(1-cos^2x)^2}{(1-sen^2x)^2}=tg^4x\\ \\ \frac{(sen^2x)^2}{(cos^2x)^2}=tg^4x\\ \\ \frac{sen^4x}{cos^4x}=tg^4x\\ \\ \boxed{tg^4x=tg^4x \ \ \ \ c.q.d}$

Answer 2

$sen^{2}x+cos^{2}x=1\\sen^{2}x-1=-cos^{2}x\\\\sen^{2}x+cos^{2}x=1\\cos^{2}x-1=-sen^{2}x$
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$1-2cos^{2}x+cos^{4}x=cos^{4}x-2cos^{2}x+1\\1-2cos^{2}x+cos^{4}x=(cos^{2}x)^{2}-2.cos^{2}x.1+1^{2}\\1-2cos^{2}x+cos^{4}x=(cos^{2}x-1)^{2}\\1-2cos^{2}x+cos^{4}x=(-sen^{2}x)^{2}\\1-2cos^{2}x+cos^{4}x=(sen^{2}x)^{2}$

$1-2sen^{2}x+sen^{4}x=sen^{4}x-2sen^{2}x+1\\1-2sen^{2}x+sen^{4}x=(sen^{2}x)^{2}-2.sen^{2}x.1+1^{2}\\1-2sen^{2}x+sen^{4}x=(sen^{2}x-1)^{2}\\1-2sen^{2}x+sen^{4}x=(-cos^{2}x)^{2}\\1-2sen^{2}x+sen^{4}x=(cos^{2}x)^{2}$
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$\frac{1-2cos^{2}x+cos^{4}x}{1-2sen^{2}x+sen^{4}x}=\frac{(sen^{2}x)^{2}}{(cos^{2}x)^{2}}\\\\\frac{1-2cos^{2}x+cos^{4}x}{1-2sen^{2}x+sen^{4}x}=(\frac{sen^{2}x}{cos^{2}x})^{2}\\\\\frac{1-2cos^{2}x+cos^{4}x}{1-2sen^{2}x+sen^{4}x}=(tg^{2}x)^{2}\\\\\boxed{\boxed{\frac{1-2cos^{2}x+cos^{4}x}{1-2sen^{2}x+sen^{4}x}=tg^{4}x}}$