The maximum value of $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 + \cos \theta & 1 & 1 \end{vmatrix}$ is (where, $\theta$ is real number)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{1}{2}$ ##
Since, $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 + \cos \theta & 1 & 1 \end{vmatrix}$
On expanding along $R_1$, we get
$\begin{aligned} \Delta &= 1 [(1 + \sin \theta) - 1] - 1 [1 - (1 + \cos \theta)] + 1 [1 - (1 + \sin \theta)(1 + \cos \theta)] \\ &= [\sin \theta] - [-\cos \theta] + [1 - 1 - \cos \theta - \sin \theta - \sin \theta \cdot \cos \theta] \\ &= \sin \theta + \cos \theta - \cos \theta - \sin \theta - \sin \theta \cdot \cos \theta \\ &= -\sin \theta \cdot \cos \theta \\ &= -\frac{1}{2} \cdot 2 \sin \theta \cdot \cos \theta \quad \text{[divide and multiply by 2]} \\ &= -\frac{1}{2} \sin 2\theta \quad [∵\sin 2\theta = 2 \sin \theta \cdot \cos \theta] \end{aligned}$
$\Delta$ is maximum when $\sin 2\theta$ is minimum.
Since, $-1 \leq \sin 2\theta \leq 1$
$∴\Delta_{\max} = -\frac{1}{2} (-1) = \frac{1}{2}$