Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:

Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$

This is what I have done 
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S.                = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y\cos y + \sin y + \cos y + 1)$
$= 2(\sin y\cos y + \sin y + \cos y + \sin^2 y + \cos^2 y)  (\sin^2 y + \cos^2 y = 1)$
$= 2(\sin^2 y + \sin y\cos y + \sin y + \cos^2 y + \cos y)$
I got stuck here. I do not know what to do from here. 
I have tried and tried several days even contacted friends but all to no avail.

If you have difficulty finding clever groupings, you could brute-force the problem by expanding everything and seeing what happens:
Writing $c := \cos y$ and $s := \sin y$ to save typing, we have
$$\require{cancel}\begin{align}
2(s+1)(c+1) &= (s+c+1)^2 \\
2(s c + s + c + 1) &= s^2 + c^2 + 1 + 2 s c + 2 s + 2 c \\
\cancel{2s c} + \cancel{2s} + \cancel{2c} + 2 &= s^2 + c^2 + 1 + \cancel{2 s c} + \cancel{2 s} + \cancel{2 c} \\
2-1 &= s^2 + c^2 \\
1 &= \sin^2 y + \cos^2 y \quad\checkmark
\end{align}$$
Since the steps are reversible, the identity is verified. $\square$
