Showing that, if $\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$, then $\sqrt2\sin\beta=\sin\alpha-\cos\alpha$ 
If
$$\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$$
then prove that
$$\sqrt2\sin\beta=\sin\alpha-\cos\alpha$$

I have been trying to solve this exercise but I don't get it. I need help. Thanks.
 A: You can use the fact that
$$
\sin^2\beta=\frac{\sin^2\beta}{\cos^2\beta+\sin^2\beta}=\frac{\tan^2\beta}{1+\tan^2\beta}
$$
and this will show that
$$
\sin^2\beta=\frac{(\sin\alpha-\cos\alpha)^2}{2}\tag{*}
$$
On the other hand, the statement is generally false. Take $\alpha=\pi/3$ and $\beta=\pi+\arctan(2-\sqrt{3})$, after having observed that
$$
\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}=2-\sqrt{3}
$$
Then $\sin\beta<0$, but $\sin\alpha-\cos\alpha>0$.
Therefore (*) is the best you can prove.
A: Writing $\alpha=x,\beta=y,$
As $(\sin x-\cos x)^2+(\sin x+\cos x)^2=2$
Using the given condition,
$2=(\sin x-\cos x)^2(1+\cot^2y)$
$\implies(\sin x-\cos x)^2=2\sin^2y$
As we have taken square, we have introduced
When do we get extraneous roots?
For example, $$a=b\implies a^2=b^2\implies a=\pm b$$
But actually $a\ne-b$
A: Avoid squaring whenever possible as it immediately introduces
When do we get extraneous roots?
$$\tan\beta=\dfrac{1-\tan\alpha}{1+\tan\alpha}=\tan(\pi/4-\alpha)$$
$$\beta=n\pi+\pi/4-\alpha$$ where $n$ is any integer
$$\implies\sin\beta=\sin(n\pi+\pi/4-\alpha)$$
Now as $\sin(n\pi+x)=(-1)^n\sin x,$
$$\sin\beta=(-1)^n\sin(\pi/4-\alpha)=?$$
A: Here's a diagrammatic verification for acute $\alpha$ and $\beta$:

