<s>[INST] 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. 
 [/INST] HINT:
$$\cos y=\sin(x-y)=\sin x\cos y-\cos x\sin y$$
$$\iff\cos x\sin y=\cos y(\sin x-1)\implies\dfrac{\sin y}{\cos y}=\dfrac{\sin x-1}{\cos x}$$
Now $\cos^2x=(1-\sin x)(1+\sin x)\implies\dfrac{1-\sin x}{\cos x}=\dfrac{\cos x}{1+\sin x}$
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