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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.

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  • $\begingroup$ That should be x on the RHS instead of y. $\endgroup$ – MonK May 28 '15 at 9:29
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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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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*}

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  • $\begingroup$ To obtain $\sin x$, $\cos x$, and $\tan x$, type \sin x, \cos x, and \tan x, respectively, in math mode. $\endgroup$ – N. F. Taussig May 28 '15 at 10:59

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