If g(x) is the inverse function and `f'(x) = sin x` then prove that `g'(x) = cosec [g(x)]`

Explanation

Since g is the inverse function of f, we have f(g(x))=x
`rArr" "(d)/(dx)(f(g(x)))=1`
`rArr" "f'(g(x)).g'(x)=1`
`rArr" "g'(x)=(1)/(sin {g(x)})`