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. `n^(2) (y^(2) - 4)`
B. `n^(2) (4 - y^(2))`
C. `n^(2) (y^(2) + 4)`
D. None of these

Answer

C

Explanation

`(dx)/(d theta) = sec theta tan theta + sin theta = tan theta (sec theta + cos theta)`
` (dy)/(dx) = n [ sec^(n-1) theta sec tan theta + cos^(n-1) theta sin theta ] `
`= n tan theta (sec^(n) theta + cos^(n) theta)`
`therefore (dy)/(dx) = (dy//d theta)/(dx//d theta) = (n tan theta (sec^(n) theta + cos^(n) theta ))/(tan theta (sec theta + cos theta ))`
` rArr ((dy)/(dx))^(2) = n^(2) . (sec^(2) theta - cos^(n) theta ^(2) + 4)/((sec theta - cos theta )^(2) + 4)`
` = (n^(2) (y^(2) +4))/((x^(2) + 4))`
` (x^(2) + 4)((dy)/(dx))^(2) = n^(2) (y^(2) + 4)`