<s>[INST] 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.
 [/INST] 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 $$(1-\sin\alpha)(1+\cos\alpha)$$
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