For all values of $θ∈(0,π/2)$ the determinant of the matrix $\begin{bmatrix}-2&\tan θ+ \sec^2θ&3\\-\sin θ&\cos θ&\sin θ\\-3&-4&3\end{bmatrix}$ always lies in the interval

$(3,5]$

Let f(θ) be the determinant of the given matrix. Then

$f(θ)=\begin{bmatrix}-2&\tan θ+ \sec^2θ&3\\-\sin θ&\cos θ&\sin θ\\-3&-4&3\end{bmatrix}$

$⇒f(θ)=\begin{bmatrix}1&\tan θ+ \sec^2θ&3\\0&\cos θ&\sin θ\\0&-4&3\end{bmatrix}$  [Applying $C_1 → C_1 +C_3$]

$⇒f(θ)= 3 \cos θ+4 \sin θ ⇒f(θ)= 5 \sin θ(θ+\tan^{-1}\frac{3}{4})$

Now, $θ∈(0,\frac{π}{2})$

$⇒θ+\tan^{-1}\frac{3}{4}∈(\tan^{-1}\frac{3}{4},\frac{π}{2}+\tan^{-1}\frac{3}{4})$

$⇒min\,f(θ)= 5\sin(\tan^{-1}\frac{3}{4})$ and, $max\,f(θ)= 5\sin\frac{π}{2}$

$⇒min\,f(θ)= 5\sin(\sin^{-1}\frac{3}{5})=3$ and, $max\,f(θ)= 5$

$⇒f(θ)∈(3,5]$