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)`

Options

A. 8
B. 16
C. 64
D. 49

Answer

C

Explanation

`(dy)/(dx)=(((dy)/(d theta)))/(((dx)/(d theta)))=(8sec^(8) theta tan theta+8cos^(7) thetasin theta)/(sec thetatan theta+sin theta)`
`=(8 tan theta(sec^(8)theta+cos^(8)theta)^(2))/(tan theta (sec theta+ cos theta))`
`therefore" "((dy)/(dx))^(2)=(64(sec^(8)theta+cos^(8)theta)^(2))/((sec theta+cos theta)^(2))`
`=(64[(sec^(8)theta-cos^(8)theta)^(2)+4])/([(sec theta-cos theta)^(2)+4])`
`=(64(y^(2)+4))/((x^(2)+4))`
`therefore" "((x^(2)+4)/(y^(2)+4))((dy)/(dx))^(2)=64`