If \(\begin{vmatrix}1&\cos\theta&0\\\sin\theta&1&\cos\theta\\\cos\theta&1&-\sin\theta\end{vmatrix} = A\sin\theta + B\cos\theta + C\sin\theta\cos\theta\) then:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(B - A = 1\)
Expanding the determinant along the first row:
\(\displaystyle \Delta = 1\begin{vmatrix}1&\cos\theta\\1&-\sin\theta\end{vmatrix} - \cos\theta\begin{vmatrix}\sin\theta&\cos\theta\\\cos\theta&-\sin\theta\end{vmatrix} + 0\)
\(= (-\sin\theta - \cos\theta) - \cos\theta(-\sin^2\theta - \cos^2\theta)\)
\(= -\sin\theta - \cos\theta + \cos\theta\)
\(= -\sin\theta\)
Comparing coefficients with \(A\sin\theta + B\cos\theta + C\sin\theta\cos\theta\):
\(A = -1,\quad B = 0,\quad C = 0\)
Checking the options:
\(A + B = -1,\quad B + C = 0,\quad C + A = -1,\quad B - A = 1\)
Only \(B - A = 1\) holds.