Determinants — Class 12 Maths Solution

Since
$\Delta = \left| {\begin{array}{cccccccccccccccccccc}1&1&1\\1&{1 + \sin \theta }&1\\{1 + \cos \theta }&1&1\end{array}} \right|$

$= \left| {\begin{array}{cccccccccccccccccccc}0&0&1\\0&{\sin \theta }&1\\{\cos \theta }&0&1\end{array}} \right|$

[ and ${C_2} \to {C_2} - {C_3}$]
$= 1(\sin \theta \cdot \cos \theta )$

$= \frac{1}{2} \cdot 2\sin \theta \cos \theta = \frac{1}{2}\sin 2\theta$

Since, the maximum value of $\sin 2\theta$ is 1.

So, for maximum value of $\theta$ should be ${45^\circ }$.

$\therefore \Delta = \frac{1}{2}\sin 2 \cdot {45^\circ }$

$= \frac{1}{2}\sin {90^\circ } = \frac{1}{2} \cdot 1 = \frac{1}{2}$