If $Δ=\begin{vmatrix}\cos α&-\sin α&1\\\sin α&\cos α&1\\\cos(α+β)&-\sin(α+β)&1\end{vmatrix}$, then

$Δ∈[1 -\sqrt{2},1+ \sqrt{2}]$

We have

$Δ=\begin{vmatrix}\cos α&-\sin α&1\\\sin α&\cos α&1\\\cos(α+β)&-\sin(α+β)&1\end{vmatrix}$

$Δ=1+ \sin β-\cos β$.

Now,

$-\sqrt{2}≤ \sin β-\cos β≤ \sqrt{2}$

$⇒1-\sqrt{2}≤1+\sin β-\cos β≤1+\sqrt{2}$

$⇒Δ∈[1 -\sqrt{2},1+ \sqrt{2}]$