If `x=cos e ctheta-sintheta` and `y=cos e c^ntheta-sin^ntheta,` then show that `(x^2+4)((dy)/(dx))^2=n^2(y^2+4)dot`

Explanation

As x = cosec theta - sin theta, we have
`x^(2)+4=(cosec theta- sin theta)^(2)+4=(cosec theta + sin theta)^(2)" (1)"`
`"and "y^(2)+4=(cosec^(n)theta - sin^(n)theta)^(2)+4 = (cosec^(n)theta+ sin^(n)theta)^(2)" (2)"`
Now,
`(dy)/(dx)=(((dy)/(d""theta)))/(((dx)/(d""theta)))=(n(cosec^(n-1)theta)(-cosec theta cot theta)-n sin^(n-1)theta cos theta)/(-cosec theta cot theta- cos theta)`
`=(n(cosec^(n)thetacot theta +sin^(n-1)theta cos theta))/((cosec theta cot theta +cos theta))`
`(ncot theta (cosec^(n)theta + sin^(n)theta))/(cot theta (cosec theta +sin theta))`
`=(n(cosec^(n)theta+sin^(n)theta))/((cosec theta +sin theta))=(nsqrt(y^(2)+4))/(sqrt(x^(2)+4))" [from (1) and (2) ]"`
`"Squaring both sides, we get "((dy)/(dx))^(2)=(n^(2)(y^(2)+4))/((x^(2)+4))`
`"or "(x^(2)+4)((dy)/(dx))^(2)=n^(2)(y^(2)+4)`