If `x=sectheta-costheta` and `y=sec^n theta- cos^n theta` then show that `(x^2+4)((dy)/(dx))^2=n^2(y^2+4)`

Explanation

`(dy)/(dx)=(((dy)/(d theta)))/(((dx)/(d theta)))=(nsec^(n) theta tan theta+ncos^(n-1) thetasin theta)/(sec thetatan theta+sin theta)`
`=(n tan theta(sec^(n)theta+cos^(n)theta)^(2))/(tan theta (sec theta+ cos theta))`
`therefore" "((dy)/(dx))^(2)=(n^2(sec^(n)theta+cos^(n)theta)^(2))/((sec theta+cos theta)^(2))`
`=(n^2[(sec^(n)theta-cos^(n)theta)^(2)+4])/([(sec theta-cos theta)^(2)+4])`
`=(n^2(y^(2)+4))/((x^(2)+4))`
`therefore" "((x^(2)+4)/(y^(2)+4))((dy)/(dx))^(2)=n^2`
`(x^2+4)((dy)/(dx))^2=n^2(y^2+4)`