How prove this $\frac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\frac{1-2\cos{x}}{1-2\sin{x}}$ let $\dfrac{\pi}{2}<x<\dfrac{3\pi}{2},0<y<\dfrac{\pi}{2}$,and such
$$\dfrac{1-\sin{x}}{1-\cos{x}}=\dfrac{1-\sin{y}}{1-\cos{y}}$$
show that
$$\dfrac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\dfrac{1-2\cos{x}}{1-2\sin{x}}$$
my idea:
$$\Longleftrightarrow (\cos{x}-\sin{y})(1-2\sin{x})=(\sin{x}-\cos{y})(1-2\cos{x})$$
But I fell this follow can't solve this problem,Thank you
 A: I'm sure that there is at least one sublimer solution than this as this is not how the problem came along , but its legitimate.
If we cross multiply the relation to be proved, we find 
$$\sin x+\sin y-(\cos x+\cos y)=-2\cos(x+y)$$
Using Prosthaphaeresis & Double Angle Formulas,
$$2\cos\frac{x-y}2\left(\sin\frac{x+y}2-\cos\frac{x+u}2\right)=-2\left(\cos^2\frac{x+u}2-\sin^2\frac{x+u}2\right)$$
Now we can safely cancel $\displaystyle\cos\frac{x+y}2-\sin\frac{x+y}2$ as for  $\displaystyle\cos\frac{x+u}2-\sin\frac{x+y}2=0\implies\tan\frac{x+y}2=1$
$\displaystyle\iff\frac{x+y}2=n\pi+\frac\pi4\iff x+y=2n\pi+\frac\pi2$ where $n$ is any integer
But, this is not possible due to the given ranges of $x,y$
So, we need $\displaystyle\cos\frac{x-y}2=\cos\frac{x+y}2+\sin\frac{x+y}2$
$\displaystyle\implies\sin\frac{x+y}2=\cos\frac{x-y}2-\cos\frac{x+y}2$
$\displaystyle\implies\sin\frac x2\cos\frac y2+\cos\frac x2\sin\frac y2=2\sin\frac x2\sin\frac y2$
$\displaystyle\implies\cot\frac x2+\cot\frac y2=2\ \ \ \ (1) $

$$\text{Now, }\frac{1-\sin x}{1-\cos x}=\frac{1-2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}=\frac12\csc^2\frac x2-\cot\frac x2$$
$$=\frac12\left(1+\cot^2\frac x2\right)-\cot\frac x2=\frac{\cot^2\dfrac x2-2\cot\dfrac x2+1}2$$
Method $\#1:$
So from the given condition, if we set $$\frac{1-\sin x}{1-\cos x}=\frac{1-\sin y}{1-\cos y}=K$$
$\displaystyle\implies\cot\frac x2,\cot\frac y2$ will be roots of 
$$\frac{\cot^2\dfrac u2-2\cot\dfrac u2+1}2=K\iff\cot^2\dfrac u2-2\cot\dfrac u2+1-2K=0$$
$\displaystyle\implies\cot\frac x2+\cot\frac y2=\frac21$ which is same as $(1)$
Method $\#2:$
So from the given condition, $\displaystyle\frac{\cot^2\dfrac x2-2\cot\dfrac x2+1}2=\frac{\cot^2\dfrac y2-2\cot\dfrac y2+1}2$
$\displaystyle\implies \left(\cot\dfrac x2-\cot\dfrac y2\right)\left(\cot\frac x2+\cot\frac y2-2\right)=0$
If $\displaystyle\cot\dfrac x2-\cot\dfrac y2=0,\tan\frac x2=\tan\frac y2\implies \dfrac x2=\dfrac y2+m\pi\iff x=2m\pi+y$  where $m$ is any integer
But, this is not possible due to the given ranges of $x,y$
