Let $Δ=\begin{vmatrix} 1 & sin \theta & 1\\-sin\theta & 1 & sin \theta \\-1 & -sin \theta & 1 \end{vmatrix}$ then Δ lies in the interval :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [2, 4]
$Δ=\begin{vmatrix} 1 & sin \theta & 1\\-sin\theta & 1 & sin \theta \\-1 & -sin \theta & 1 \end{vmatrix}$
$=1(1+\sin θ)+\sin θ(-\sin θ+\sin θ)+1(1+\sin^2 θ)$
$=2+2\sin^2 θ$
as $0≤\sin^2 θ≤1⇒1≤1+\sin^2 θ≤2$
so $2≤2(1+\sin^2 θ)≤4$
so $Δ∈[2, 4]$