How would you solve the trigonometric identity (cosx-sinx)/(cosx+sinx)=sec2x-tan2x?

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1. sec 2x = 1/cos2x = 1/(cos^2x - sin^2x)
tan 2x = sin 2x / cos 2x = sin 2x / (cos^2 x - sin^2 x)
so
sec 2x -tan 2x = (1 - sin 2x) / (cos^2 x - sin^2 x)
= (1 - 2 sin x cos x) / (cos^2 x - sin^2 x)
= (1 - 2 sin x cos x)/ [(cos x - sin x)(cos x + sin x) ]
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now the left side
(cos x - sin x)(cos x - sin x) / [ cos x + sin x)(cos x - sin x)]
(cos^2 x - 2 sin x cos x + sin^2 x) / [ cos x + sin x)(cos x - sin x)]
(1 - 2 sin x cos x) / [ cos x + sin x)(cos x - sin x)]
remarkable, same as the right side

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👨‍🏫
Damon

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