# 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.

• That should be x on the RHS instead of y. – MonK May 28 '15 at 9:29

$$\cos y=\sin(x-y)=\sin x\cos y-\cos x\sin y$$
$$\iff\cos x\sin y=\cos y(\sin x-1)\implies\dfrac{\sin y}{\cos y}=\dfrac{\sin x-1}{\cos x}$$
Now $\cos^2x=(1-\sin x)(1+\sin x)\implies\dfrac{1-\sin x}{\cos x}=\dfrac{\cos x}{1+\sin x}$
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
• To obtain $\sin x$, $\cos x$, and $\tan x$, type \sin x, \cos x, and \tan x, respectively, in math mode. – N. F. Taussig May 28 '15 at 10:59