Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$? Why  $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?
I am learning trigonometric identities one identity I have to proof is the next:
$ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$
so I tried to resolve the identity for the left:
$  1 + \sin^2 \alpha+ \cos^2\alpha - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha  $
$=  1 + 1 - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha  $ 
$=  2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
And I got stuck, I did not know what to do, so I went to see the problem's answer and I was going fine, the part I was not able to resolve is the next one:
$  2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$ 
$=  2 (1  + \cos\alpha - \sin\alpha(1+\cos\alpha))$
$=  2 (1 - \sin \alpha)(1+ \cos \alpha)$
So the question is how did the teacher do the last three steps? I  cant figure it out.
 A: They are just the distributive property

$$a(b+c)=ab+ac$$

in the first one we have $a=-\sin\alpha, b= 1, c=\cos\alpha$ which comes from the terms
$$2(1+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{1}_{b}+\cos\alpha+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{\cos\alpha}_{c})$$
The next step is the same property with $a=(1+\sin\alpha), b=1, c=\cos\alpha$ as seen
$$2(\underbrace{(1+\cos\alpha)}_{a}\cdot \underbrace{1}_{b}+\underbrace{(1+\cos\alpha)}_{a}\underbrace{(1-\sin\alpha)}_{c}).$$
A: You calculated that the left side is equal to $$2(1-\sin\alpha + \cos\alpha -\sin\alpha\cos\alpha).$$
Now, try to prove that the left side is equal to that as well. Try expanding $$(1-\sin\alpha)(1+\cos\alpha)$$
A: Indeed
$$
(1 - \sin \alpha + \cos \alpha)^2 = (1 - \sin \alpha)^2 + \cos^2 \alpha+ 2(1 - \sin \alpha)\cos \alpha 
$$
$$
= (1 - 2\sin \alpha + \sin^2\alpha + \cos^2\alpha) +  2(1 - \sin \alpha)\cos \alpha 
$$
$$
= 2(1 - \sin \alpha) + 2(1 - \sin \alpha)\cos \alpha = 2(1 - \sin \alpha)(1 + \cos \alpha)
$$
