trigonometry solution $$\frac{\sin A - \cos A}{\sin A+\cos A}=\frac{X}{Y}$$
Then prove that $X^2 + Y^2 = 2$.
Answer :
$\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{x}{y}$
$\implies x=k(\sin A-\cos A)$ and 
         $y=k(\sin A+\cos A)$. 

Squaring on both sides and adding 

$x^2+y^2=k^2(\sin A-\cos A)^2+k^2(\sin A+\cos A)^2$

$\implies x^2+y^2=k^2[\sin^2A+\cos^2A-2\sin A\cos A+\sin^2A+\cos^2A+2\sin A\cos A]$

$\implies x^2+y^2=2 k^2$
But how can eliminate k.means by what logic we can say k=+1 or -1.
Please suggest. Thanks.
It is impossible to prove that result, mainly because it is not correct. Take $A = 0$ then $$\frac{\sin A - \cos A}{\sin A + \cos A} = -1$$ now take $X = k$ for some $k \neq \pm1$ and take $Y = -k$ so that $$\frac{X}{Y} = -1$$ but since $k \neq \pm1$ then $$X^2 + Y^2 = 2k^2 \neq 2.$$
You can simply substitute $X$ with $(\sin A - \cos A)$ and $Y$ with $(\sin A + \cos A)$. If we do this we see that the first equality is obviously satisfied. The $k$ term that you multiply this by is unnecessary. You can see how I proved this here.