How do you prove sec^2x-2secxcosx+cos^2x=tan^2x-sin^2xsec2x−2secxcosx+cos2x=tan2x−sin2x ?

Prove:
sec^2x-2secxcosx+cos^2x=tan^2x-sin^2xsec2x−2secxcosx+cos2x=tan2x−sin2x

I boiled it down to each side being 1 but I don't think I did it all the way correctly.

3 Answers
Jim H
Nov 26, 2017

Please see below.

Explanation:

secxcosx=1secxcosx=1, and sec^2x-1 = tan^2xsec2x−1=tan2x and 1-cos^2x =sin^2x1−cos2x=sin2x

So we have

sec^2x-2secxcosx+cos^2x = sec^2x-2+cos^2xsec2x−2secxcosx+cos2x=sec2x−2+cos2x

= sec^2x-1-1+cos^2x=sec2x−1−1+cos2x

= (sec^2x-1)-(1-cos^2x)=(sec2x−1)−(1−cos2x)

= tan^2x-sin^2x=tan2x−sin2x

Sean
Nov 26, 2017

Please see below.

Explanation:

.

We know:

sec^2x=tan^2x+1sec2x=tan2x+1 and

secx=1/cosxsecx=1cosx

Let's plug it into the left side:

=tan^2x+1-2(1/cosx)cosx+cos^2x=tan2x+1−2(1cosx)cosx+cos2x

=tan^2x+1-2(1/cancelcolor(red)cosx)cancelcolor(red)cosx+cos^2x

=tan^2x+1-2+cos^2x

But from the identity sin^2x+cos^2x=1 we can get:

cos^2x=1-sin^2x. Let's plug this in:

=tan^2x+1-2+1-sin^2x

=tan^2x+cancelcolor(red)1-cancelcolor(red)2+cancelcolor(red)1-sin^2x

=tan^2x-sin^2x

Jim H
Nov 26, 2017

Here is another fun way to prove it.

Explanation:

sec^2x-2secxcosx+cos^2x = (secx-cosx)^2

= (1/cosx-cosx)^2

= (1-cos^2x)^2/cos^2x

= ((1-cos^2x)(1-cos^2x))/cos^2x

= (sin^2x(1-cos^2x))/cos^2x

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

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

= tan^2x-sin^2x

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