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Let A=|1sintheta1-sintheta1sintheta-1-si...

Let `A=|1sintheta1-sintheta1sintheta-1-sintheta1|,` where `0lt=thetalt=2pidot` Then `D e t(A)=0` (b) `D e t(A) in (2,oo)` `D e t(A) in (2,4)` (d) `D e t(A) in [2,4]`

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To solve the problem, we need to find the determinant of the matrix \( A \) given by: \[ A = \begin{vmatrix} 1 & \sin \theta \\ 1 - \sin \theta & -1 \end{vmatrix} \] ...
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