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dalendrk
[tex]We\ know:\\\\\tan x=\dfrac{\sin x}{\cos x}\\\\\sec x=\dfrac{1}{\cos x}\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x\\-----------------------\\\\-\tan^2x+\sec^2x=-\dfrac{\sin^2x}{\cos^2x}+\dfrac{1}{\cos^2x}=\dfrac{-\sin^2x+1}{\cos^2x}=\dfrac{-(1-\sin^2x)}{\cos^2x}\\\\=\dfrac{-\cos^2x}{\cos^2x}=-1[/tex]

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