Question #cc719

1 Answer
Richard
Jul 24, 2016

#sin^2 2x(cot^2x-tan^2x)=4cos2x#
#=sin2x*sin2x(cot^2x-tan^2x)#

Using the double angle formula for sin and converting cot and tan to terms with sin and cos we now have

#=4sin^2xcos^2x(cos^2x/sin^2x - sin^2x/cos^2x)#
#=4sin^2xcos^2x((cos^4x-sin^4x) / (sin^2xcos^2x))#
#=4sin^2xcos^2x((cos^2x +sin^2x)(cos^2x-sin^2x))/((sin^2xcos^2x))#
#=4sin^2xcos^2x((cos^2x-sin^2x)/(sin^2x*cos^2x))# since #cos^2x +sin^2x =1 #
#=(4cos^4sin^2x - sin^4xcos^2x)/(sin^2xcos^2x)#
#=(4cancel(sin^2xcos^2x)(cos^2x-sin^2x))/(cancel(sin^2xcos^2x))#
#=4(cos^2x-sin^2x)#
#=4cos2x# since #cos^2x-sin^2x=cos2x#

Related questions
See all questions in Proving Identities
Impact of this question
1681 views around the world
You can reuse this answer
Creative Commons License