`FindIntegerNullVector` is your friend. Notice that this method allows you to obtain an even simpler expression.

**A. Find relations between sets of different functions**


    v[1]={ArcCsc[18/Sqrt[50+17 Sqrt[2]]],ArcTan[Sqrt[2]],ArcCot[3+Sqrt[2]]};
    v[2]={ArcCot[Sqrt[2]],ArcCot[2 Sqrt[2]],ArcCot[3+Sqrt[2]]};
    v[3]={ArcCot[2 Sqrt[2]],ArcCot[Sqrt[2]],ArcTan[Sqrt[2]]};
    v[4]={ArcCot[Sqrt[2]],ArcTan[Sqrt[2]],π/2};

**B. This is the main part**

    Table[FindIntegerNullVector[N[v[i],1000]],{i,4}]
    Out[1]= {{-3,-1,11},{-2,1,4},{1,1,-1},{1,1,-1}}

**C. Make a list of replacement rules**

    r[1]={ArcCsc[18/Sqrt[50+17 Sqrt[2]]]->(-ArcTan[Sqrt[2]]+11ArcCot[3+Sqrt[2]])/3};
    r[2]={ArcCot[3+Sqrt[2]]->1/4(2ArcCot[Sqrt[2]]-ArcCot[2 Sqrt[2]])};
    r[3]={ArcCot[2 Sqrt[2]]->ArcTan[Sqrt[2]]-ArcCot[Sqrt[2]]};
    r[4]={ArcCot[Sqrt[2]]->π/2-ArcTan[Sqrt[2]]};

**D. Apply the rules**

    sx=Fold[ReplaceAll,x,Array[r,4]]//Expand;
    sy=Fold[ReplaceAll,y,Array[r,4]]//Expand;
    sx-sy
    Out[2]= 0

Comment/observation
-------------------

In principle, all but one identities found by the LLL method are easy to prove by hand. Only the first one, namely 
$$11\,\mathrm{arccot}\big(3+\sqrt{2}\big) = \mathrm{arctan}\big(\sqrt{2}\big)+ 3\,\mathrm{arccsc}\!\Bigg\{\frac{18}{\sqrt{50+17 \sqrt{2}}}\Bigg\}$$
is somewhat unexpected.