Question

Mostre que:

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

Answer 1

Olá!

$\mathsf{\dfrac{1 - 2 \cdot \cos^2 x + \cos^4 x}{1 - 2 \cdot \sin^2 x + \sin^4 x} =}$

Fatorando,

$\mathsf{\dfrac{(1 - \cos^2 x)^2}{(1 - \sin^2 x)^2} =}$


 Ademais, sabemos que: $\mathbf{\sin^2 x + \cos^2 x = 1}$. Com efeito,

$\mathbf{\sin^2 x = 1 - \cos^2 x}$

Ou,

$\mathbf{\cos^2 x = 1 - \sin^2 x}$


 Substituindo,

$\\ \mathsf{\dfrac{(1 - \cos^2 x)^2}{(1 - \sin^2 x)^2} =} \\\\\\ \mathsf{\dfrac{(\sin^2 x)^2}{(\cos^2 x)^2} =} \\\\\\ \mathsf{\dfrac{\sin^4 x}{\cos^4 x} =} \\\\\\ \mathsf{\left(\dfrac{\sin x}{\cos x}\right)^4=} \\\\\\ \boxed{\mathsf{\tan^4 x}}$

Answer 2

$\mathsf{\dfrac{1 - 2.\cos^2 x + \cos^4 x}{1 - 2.\sin^2 x + \sin^4 x} =}$

$\mathsf{\dfrac{(1 - \cos^2 x)^2}{(1 - \sin^2 x)^2} =}$

$\\ \mathsf{\dfrac{(1 - \cos^2 x)^2}{(1 - \sin^2 x)^2} =} \\\mathsf{\dfrac{(\sin^2 x)^2}{(\cos^2 x)^2} =} \\\mathsf{\dfrac{\sin^4 x}{\cos^4 x} =} \\\mathsf{\left(\dfrac{\sin x}{\cos x}\right)^4=} \\\huge\boxed{\boxed{\boxed{\mathsf{\tan^4 x}}}}}$