How do you prove the identity (sinx-cosx)/cos^2x=(tan^2x-1)/(sinx+cosx)?

1 Answer
George C.
Sep 30, 2016

See explanation...

Explanation:

The difference of squares identity can be written:

a^2-b^2 = (a-b)(a+b)

Use this with a = sin x and b = cos x to find:

(sin x - cos x)/(cos^2 x) = ((sin x - cos x)(sin x + cos x))/((cos^2 x)(sin x + cos x)

color(white)((sin x - cos x)/(cos^2 x)) = (sin^2 x - cos^2 x)/((cos^2 x)(sin x + cos x)

color(white)((sin x - cos x)/(cos^2 x)) = ((sin^2 x)/(cos^2 x) - (cos^2 x)/(cos^2 x))/(sin x + cos x)

color(white)((sin x - cos x)/(cos^2 x)) = (tan^2 x - 1)/(sin x + cos x)

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