Question

Question #de16d

Answer 1

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`

Answer 2

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`