Trigonometry Identity proving If $\sin(x-y) =\cos y$ prove that $\tan y = \frac{1+ \sin y}{\cos y} $. 
Is there an error with the question? I don't seem to be able to get the answer. Should it be $\tan x$ instead of $\tan y = \frac{1+\sin y}{\cos y}$ ?
$$\tan x\times \cos y = 1 + \sin y $$
$$\frac{\sin x}{\cos x} \times \cos y = \frac{\sin x\times\cos y}{\cos x}$$
$$\frac{\sin x\times\cos y}{\cos x}= 1 + \sin y $$
$$\frac{\sin x\times\cos y}{\sin(x-y)}= 1 + \sin y $$
I am stuck at this step. 

Since $\sin(x-y)= \cos(y)$, expanding it gives
\begin{equation*}
\sin (x) \cos(y)-\cos(x) \sin(y)=\cos(y).
\end{equation*}
Dividing both sides by $\cos(y)$ we have
\begin{equation*}
\sin(x)-\cos(x)\tan(y)=1.
\end{equation*}
So from here, we get
\begin{equation*}
\tan(y)=\dfrac{\sin(x)-1}{\cos(x)}.
\end{equation*}
