How does one verify #(cos^2x-sin^2x)/(1-tan^2x)=cos^2x#?

Question

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

I know that there are a few properties in here, but I don't know how the signs would effect them.

9068 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T03:41:41

Image reference...

, my notebook...

Answer 2 · 2018-04-09T03:46:54

The given equation is,
#(cos^2x−sin^2x)/(1−tan^2x)=cos^2x#

LHS.

#=>(cos^2x−sin^2x)/(1−sin^2x/cos^2x)#

#=>(cos^2x−sin^2x)/((cos^2x−sin^2x)/cos^2x)#

#=>(cancel((cos^2x−sin^2x))cos^2x)/cancel((cos^2x−sin^2x)#

#=> cos^2x#

Hence Proved.! :)