Respuesta :

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[tex]2\sec^2x-2\sec^2x\sin^2x-\sin^2x-\cos^2x=2\sec^2x(1-\sin^2x)-(\sin^2x+\cos^2x)[/tex]

Since [tex]\sin^2x+\cos^2x=1[/tex], you have

[tex]2\sec^2x(1-\sin^2x)-(\sin^2x+\cos^2x)=2\sec^2x\cos^2x-1[/tex]

and since [tex]\sec x=\dfrac1{\cos x}[/tex], you end up with [tex]2-1=1[/tex].

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