Respuesta:
Queda demostrado!
Explicación paso a paso:
■》Solución:
[tex]sec^{4}x-sec^{2}x=tan^{4}x + tan^{2}x \\ \\ sec^{2}x( sec^{2}x - 1) = tan^{2}x(tan^{2}x + 1) \\ \\ \frac{sec^{2}x}{tan^{2}x} = \frac{tan^{2}x + 1}{sec^{2}x - 1} \\ \\ \frac{ \frac{1}{cos^{2}x} }{ \frac{sen^{2}x}{cos^{2}x}} =\frac{ \frac{sen^{2}x + cos^{2}x}{cos^{2}x} }{ \frac{1 -cos^{2}x }{cos^{2}x} } \\ \\ \frac{1}{sen^{2}x} = \frac{1}{1 -cos^{2}x} \\ \\ 1 - cos^{2}x = sen^{2}x \\ \\ 1 = sen^{2}x + cos^{2}x \\ \\ 1 = 1 \\ \\ queda \: demostrado...[/tex]
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Verified answer
Hola!
Respuesta:
Queda demostrado!
Explicación paso a paso:
■》Solución:
[tex]sec^{4}x-sec^{2}x=tan^{4}x + tan^{2}x \\ \\ sec^{2}x( sec^{2}x - 1) = tan^{2}x(tan^{2}x + 1) \\ \\ \frac{sec^{2}x}{tan^{2}x} = \frac{tan^{2}x + 1}{sec^{2}x - 1} \\ \\ \frac{ \frac{1}{cos^{2}x} }{ \frac{sen^{2}x}{cos^{2}x}} =\frac{ \frac{sen^{2}x + cos^{2}x}{cos^{2}x} }{ \frac{1 -cos^{2}x }{cos^{2}x} } \\ \\ \frac{1}{sen^{2}x} = \frac{1}{1 -cos^{2}x} \\ \\ 1 - cos^{2}x = sen^{2}x \\ \\ 1 = sen^{2}x + cos^{2}x \\ \\ 1 = 1 \\ \\ queda \: demostrado...[/tex]