How prove this $\frac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\frac{1-2\cos{x}}{1-2\sin{x}}$

Question

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

Answer 1

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) $

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$$\text{Now, }\frac{1-\sin x}{1-\cos x}=\frac{1-2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}=\frac12\csc^2\frac x2-\cot\frac x2$$

$$=\frac12\left(1+\cot^2\frac x2\right)-\cot\frac x2=\frac{\cot^2\dfrac x2-2\cot\dfrac x2+1}2$$

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 $x,y$

Answer 2

Let us use the duplication formulas

$$ \cos(\theta)=\frac{1-t^2}{1+t^2}, \ \sin(\theta)=\frac{2t}{1+t^2}, \ \text{where} \ t=\tan\big(\frac{\theta}{2}\big) \tag{1} $$

We deduce

$$ \frac{1-\sin(\theta)}{1-\cos(\theta)}= \frac{1-\frac{2t}{1+t^2}}{1-\frac{1-t^2}{1+t^2}}= \frac{(t-1)^2}{2t^2} \tag{2} $$

Let us then put $t_x=\tan\big(\frac{x}{2}\big)$ and $t_y=\tan\big(\frac{y}{2}\big)$. Since $\cos(x)<0$ and $\cos(y)>0$, we have $\cos(x)\neq\cos(y)$ and hence $t_x\neq t_y$. By (2) we have $\frac{(t_x-1)^2}{2t_x^2}=\frac{(t_y-1)^2}{2t_y^2}$, or in other words

$$ \begin{array}{lcl} 0 &=& (t_x-1)^2t_y^2-(t_y-1)^2t_x^2 \\ &=& \bigg((t_x-1)t_y+(t_y-1)t_x\bigg)\bigg((t_x-1)t_y-(t_y-1)t_x\bigg) \\ &=& \bigg(2t_xt_y-t_x-t_y\bigg)\bigg(t_y-t_x\bigg) \\ &=& 2t_xt_y-t_x-t_y \ (\text{because} \ t_x\neq t_y). \end{array}\tag{3} $$

Next, expanding $\Delta=(\cos(x)-\sin(y))(1-2\sin(x))-(\sin(x)-\cos(y))(1-2\cos(x))$ we see that

$$ \Delta= \cos(x)-\sin(x)+\cos(y)-\sin(y)-2\cos(x)\cos(y)+2\sin(x)\sin(y) $$

Let us put $A=\cos(x)-\sin(x)+\cos(y)-\sin(y)$ and $B=\cos(x)\cos(y)-\sin(x)\sin(y)$.

We have

$$ \begin{array}{lcl} A&=& \frac{1-2t_x-t_x^2}{1+t_x^2}+ \frac{1-2t_y-t_y^2}{1+t_y^2} \\ &=& \frac{1-2t_x-t_x^2+t_y^2-2t_xt_y^2-t_x^2t_y^2}{(1+t_x^2)(1+t_y^2)}+ \frac{1-2t_y-t_y^2+t_x^2-2t_yt_x^2-t_x^2t_y^2}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{2-2(t_x+t_y)-2t_xt_y(t_x+t_y)-2t_x^2t_y^2}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{2-2(t_x+t_y)-(t_x+t_y)^2-\frac{1}{2}(t_x+t_y)^2}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{2-2(t_x+t_y)-\frac{3}{2}(t_x+t_y)^2}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{2-2(t_x+t_y)-\frac{3}{2}(t_x^2+t_y^2)-\frac{3}{2}(t_x+t_y)}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{2-\frac{7}{2}(t_x+t_y)-\frac{3}{2}(t_x^2+t_y^2)}{(1+t_x^2)(1+t_y^2)} \\ \end{array} $$

and

$$ \begin{array}{lcl} B&=& \frac{1-t_x^2}{1+t_x^2}\frac{1-t_y^2}{1+t_y^2} -\frac{2t_x}{1+t_x^2} \frac{2t_y}{1+t_y^2} \\ &=& \frac{1-t_x^2-t_y^2+t_x^2t_y^2}{(1+t_x^2)(1+t_y^2)} -\frac{2(t_x+t_y)}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{1-t_x^2-t_y^2+\big(\frac{t_x+t_y}{2}\big)^2}{(1+t_x^2)(1+t_y^2)}- \frac{2(t_x+t_y)}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{1-\frac{3}{4}\big(t_x^2+t_y^2\big)+\frac{t_xt_y}{2}}{(1+t_x^2)(1+t_y^2)}- \frac{2(t_x+t_y)}{(1+t_x^2)(1+t_y^2)} \\ &=& \frac{1-\frac{7}{4}\big(t_x+t_y\big)-\frac{3}{4}\big(t_x^2+t_y^2\big)}{(1+t_x^2)(1+t_y^2)} \\ \end{array} $$

We see that $A=2B$, so $\Delta=0$ as wished.