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.

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.
