[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].
