The maximum value of $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{array}\right|$ is ( $\theta$ is real)
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\sqrt{2}$
(d) $-\frac{\sqrt{3}}{2}$
$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{array}\right|$
$=\left|\begin{array}{ccc}1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ \cos \theta & 0 & 0\end{array}\right|$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]
$=-\sin \theta \cos \theta$
$=-\frac{\sin 2 \theta}{2}$
Now, Maximum and minimum value of $\sin \theta$ is 1 and $-1$.
So, the maximum value of $-\sin \theta$ is 1 .
So, the maximum value of $-\sin 2 \theta$ is 1 .
Therefore, the maximum value of $-\frac{\sin 2 \theta}{2}$ is $\frac{1}{2}$.
Hence, the correct option is (a).