2sin^2x-1=1-2/sec^2× What is the iguality of the equation?

1 Answer
Apr 24, 2018

Starting from #sin^2 x - cos^2 x = sin^2x - cos ^2x# and combining that with #cos^2 x+sin^2x=1# and # sec x=1/cos x# gives the desired result.

Explanation:

I think we're being asked to show

# 2 sin ^2 x - 1 = 1 - 2/ sec^2 x #

I can see both sides will turn out to be #-cos(2x)#. That doesn't really matter, but it gives us a place to start the proof.

#sin^2 x - cos^2 x = sin^2x - cos ^2x.#

Obviously a thing is equal to itself. Given #cos ^2 x + sin^2 x= 1# we can substitute #sin^2 x = 1 - cos^2x# or #cos^2x = 1 - sin ^2 x.# Let's do both.

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

# 3 sin^2 x - 1 = 1 - 2 cos^2 x #

# 2 sin ^2 x - 1 = 1 - 2/sec^2 x quad sqrt #