Find dy/dx by using implicit differentiation: sinx=x(1+tany)

Here's what I have.

dy/dx[sinx]=dy/dx[x(1+tany)]

cosx=x[(sec^2y)dy/dx]+(1+tany)

cosx-tany-1=dy/dx[x(sec^2y)]

dy/dx=(cosx-tany-1)/(xsec^2y)

I haven't done this in a while so I am not confident in my answer. Did I do it right?