Number of solutions of the equation $\begin{vmatrix}-1 & 0 & sin \theta \\sin \theta & -1 & 0\\0 & sin \theta & -1 \end{vmatrix}=0 $ in $(0, \pi )$ is :

exactly one

The correct answer is Option (1) → exactly one

$Δ\begin{vmatrix}-1 & 0 &\sin \theta \\\sin \theta & -1 & 0\\0 &\sin \theta & -1 \end{vmatrix}=0$

$=-1(1)+0+\sin θ(\sin^2θ)=0$

$\sin^3θ=1⇒\sin θ=1$

in (0, π) it has only one solution at $θ=\frac{π}{2}$