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.
 A: $$\sin y=s,\cos y=c$$
$$(c+s+1)^2=c^2+s^2+1+2c+2s+2cs=2\underbrace{(1+c)}+2s\underbrace{(1+c)}=?$$
Another way
As $s^2+c^2=1$
$$(s+(c+1))^2=s^2+(c+1)^2+2s(c+1)=1-c^2+(c+1)^2+2s(c+1)=(c+1)[1-c+c+1+2s]=?$$
A: Expanding the RHS,
$$\color{blue}{(\sin y + \cos y + 1 )^2} = \sin^2 y + \cos^2 y +  1 +2\cos y \sin y + 2\cos y + 2\sin y$$
$$= 1+ 1 +2\cos y \sin y + 2\cos y + 2\sin y=2(1+\cos y \sin y + \cos y + \sin y) = 2 \left[(1+\cos y) + (\sin y + \cos y \sin y ) \right] = 2 \left[(1+\cos y) + \sin y(1 + \cos y) \right]=\color{blue}{2(1+\cos y)(1+\sin y )}$$
A: 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. $\square$
A: $$RHS - LHS=(\sin y + \cos y + 1)^2-2(\sin y + 1)(\cos y + 1) $$
$$=[(\sin y +1)+( \cos y + 1) -1]^2-2(\sin y + 1)(\cos y + 1) $$
$$=(\sin y +1)^2+( \cos y + 1)^2 + 1 -2(\sin y + 1)-2(\cos y + 1) $$
$$=[(\sin y +1)^2-2(\sin y + 1)+1]+[( \cos y + 1)^2 - 2(\cos y + 1)+1] -1 $$
$$=\sin^2 y+\cos^2 y -1  = 0$$
