Question

Question #cc719

Answer 1

`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`