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

`frac{dy}{d theta}=n sec^(n-1). sec theta. tan theta-n.cos^theta(n-1)(-sin theta)`

`n tan theta(sec^ n theta+cos theta)`

`frac{dx}{d theta}=frac{sin theta}{cos theta}(sec theta+cos theta)=tan theta(sec theta+cos theta)`

`:. frac{dy}{dx}=frac{n tan theta(sec^n theta+cos^n theta)}{tan theta(sec theta+cos theta)}`

`(frac{dy}{dx})^2=frac{n^2(sec^n theta+cos^n theta)^2}{(sec theta+cos theta)^2}`

`=frac{n^2(sec^n theta-cos^n theta)^2+4}{(sec theta-cos theta)^2+4}=frac{n^2(y^2+4)}{x^2+4}`

`(x^2+4)((dy)/(dx))^2=n^2(y^2+4)`