# Prove the following identities sec(x) - sin(x) = [tan^2(x) + cos^2(x)] / [sec(x) + sin(x)]?

sec(x) - sin(x) = [tan^2(x) + cos^2(x)] / [sec(x) + sin(x)]

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Lv 4

sec(x) - sin(x) = [tan^2(x) + cos^2(x)] / [sec(x) + sin(x)]

LS:

sec(x) - sin(x)

= [sec(x) - sin(x)]*[sec(x) + sin(x)] / [sec(x) + sin(x)]

=[sec^2(x) - sin^2(x)] / [sec(x) + sin(x)]

now must show:

sec^2(x) - sin^2(x) = tan^2(x) + cos^2(x)

LS:

sec^2(x) - sin^2(x)

=sec^2(x) - 1 + cos^2(x)

aside:

sin^2(x) + cos^2(x) = 1 dividing through by cos^2(x)

tan^2(x) + 1 = sec^2(x)

tan^2(x) = sec^2(x) - 1

resubstituting tan^2(x):

=sec^2(x) - 1 + cos^2(x)

= tan^2(x) + cos^2(x) as required.

R. H. S.=[tan^2 x + cos^2 x] / [sec x + sin x]

=[sec^2 x - 1 +cos^2 x] / [sec x + sin x] [since tan^2x=sec^2x - 1]

=[sec^2 x - (1 - cos^2 x)] / [sec x + sin x]

=[sec^2 x - sin^2 x] / [sec x+ sin x]

=[sec x + sin x][sec x - sin x] / [sec x + sin x] [since a^2 - b^2=(a+b)(a-b)]

=sec x - sin x [cancelling (sec x + sin x) in both numerator n denominator]

=L. H. S.