How does one verify #(secx-cosx)^2=tan^2x-sin^2x#?

#(secx-cosx)^2=tan^2x-sin^2x#

1 Answer
Apr 23, 2018

See below.

Explanation:

We'll leave the right side alone and attempt to get the left side to match it.

Expand the left side:

#(secx-cosx)(secx-cosx)=tan^2x-sin^2x#

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

#2secxcosx=2/cancelcosxcancelcosx=2,# so we get

#sec^2x-2+cos^2x=tan^2x-sin^2x#

Rewrite, splitting up the #-2# into #-1-1#:

#(sec^2x-1)+(cos^2x-1)=tan^2x-sin^2x#

Recall the following identities:

#sec^2x-1=tan^2x#
#sin^2x+cos^2x=1 ->cos^2x-1=-sin^2x#

So, we get

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