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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