If `cos y= x cos (a+y)`, with `cos a ne pm 1`, prove that `(dy)/(dx)=(cos^(2)(a+y))/(sin a)`.

Explanation

Given relation is cos y = x cos (a+y). Therefore,
`x=(cos y)/(cos (a+y))`
Differentaiting w.r.t.y, we get
`(dx)/(dy)=(d)/(dy)((cos y)/(cos (a+y)))`
`=((cos (a+y)(-sin y)- cos y (-sin (a+y)))/(cos^(2)(a+y)))`
`=((-cos (a+y) sin y + cos y sin (a+y))/(cos^(2) (a+y)))`
`=((sin (a+y-y))/(cos^(2)(a+y)))=(sin a)/(cos^(2)(a+y))`
`therefore" "(dy)/(dx)=(cos^(2)(a+y))/(sin a)`