If $Δ=\begin{vmatrix}\cos α&-\sin α&1\\\sin α&\cos α&1\\\cos(α+β)&-\sin(α+β)&1\end{vmatrix}$, then
Answer & explanation
Correct answer: option 1
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,
$-\sqrt{2}≤ \sin β-\cos β≤ \sqrt{2}$
$⇒1-\sqrt{2}≤1+\sin β-\cos β≤1+\sqrt{2}$
$⇒Δ∈[1 -\sqrt{2},1+ \sqrt{2}]$