Question #de16d

2 Answers
Aug 6, 2016

We have: (cos^2x-sin^2x)/(cot^2x-tan^2x)

Start by writing tan and cot in terms of sin and cos

=(cos^2x-sin^2x)/((cos^2x)/(sin^2x)-sin^2x/cos^2x)

=(cos^2x-sin^2x)/((cos^4x-sin^4x)/(sin^2xcos^2x))*(sin^2xcos^2x)/(sin^2xcos^2x)

=(cos^4xsin^2x-sin^4xcos^2x)/(cos^4x-sin^4x)

=((sin^2xcos^2x)cancel((cos^2x-sin^2x)))/((cos^2x+sin^2x)cancel((cos^2x-sin^2x)))

=(sin^2xcos^2x)/1

=sin^2xcos^2x

Aug 6, 2016

We have:
LHS=(cos^2x-sin^2x)/(cot^2x-tan^2
=(sin^2xcos^2x(cos^2x/(sin^2xcos^2x)-sin^2x/(sin^2xcos^2x)))/(csc^2x-1-sec^2x+1)

=((sin^2xcos^2x)(csc^2x-sec^2x))/(csc^2x-sec^2x)

=sin^2xcos^2x