How do you prove #sec^2x-2secxcosx+cos^2x=tan^2x-sin^2x# ?

Prove:
#sec^2x-2secxcosx+cos^2x=tan^2x-sin^2x#

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=1#, and #sec^2x-1 = tan^2x# and #1-cos^2x =sin^2x#

So we have

#sec^2x-2secxcosx+cos^2x = sec^2x-2+cos^2x#

# = sec^2x-1-1+cos^2x#

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

# = tan^2x-sin^2x#

Sean
Nov 26, 2017

Please see below.

Explanation:

.

We know:

#sec^2x=tan^2x+1# and

#secx=1/cosx#

Let's plug it into the left side:

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

#=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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