Implicit differentiation of $\sin(x)=x(1+\tan(y))$

Implicit differentiation of $\sin(x)=x(1+\tan(y))$

I do most of my studying independently either before I take the class to
get ahead or after in order to fix trouble areas. Right now I'm trying to
review Single Variable Calculus. Anyway, I ran into a road block this
weekend.
Problem
Find $dy/dx$ through implicit differentiation: $$ \sin (x)=x(1+\tan(y)) $$
My solution
$$\begin{align} \frac{d}{dx}[\sin
(x)]&=\frac{d}{dx}[x(1+\tan(y))]\tag{1}\\ \cos(x)&=(1)(1+
\tan(y))+x(1+\tan(y))^{-1}(\sec^{2}(y))\frac{dy}{dx}\tag{2}\\ \cos(x)&=(1+
\tan(y))+\frac{x(\sec^{2}(y))}{1+\tan(y)}\frac{dy}{dx}\tag{3}\\
\cos(x)-(1+
\tan(y))&=\frac{x(\sec^{2}(y))}{1+\tan(y)}\frac{dy}{dx}\tag{4}\\
\frac{dy}{dx}&=\frac{(\cos(x)-1-
\tan(y))(1+\tan(y))}{x(\sec^{2}(y))}\tag{5} \end{align}$$
Solution from manual I'm using
$$\begin{align} \sin (x)&=x(1+\tan(y))\tag{6}\\ \cos(x) &= x(sec^{2}(y))y'
+ (1+\tan(y))(1)\tag{7}\\
y'&=\frac{\cos(x)-\tan(y)-1}{x\sec^{2}(y)}\tag{8} \end{align}$$
The disagreement seems to lie with steps 2 & 7. Any help figuring out why
this disagreement exists would be good. Thank you for your help.