Question #cc719

1 Answer
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