The maximum value of $\left| {\begin{array}{cccccccccccccccccccc}1&1&1\\1&{1 + \sin \theta }&1\\1&1&{1 + \cos \theta }\end{array}} \right|$ is $\frac{1}{2}$.
Correct Answer True
The maximum value of $\left| {\begin{array}{cccccccccccccccccccc}1&1&1\\1&{1 + \sin \theta }&1\\1&1&{1 + \cos \theta }\end{array}} \right|$ is $\frac{1}{2}$.
Correct Answer True
Since, $\left| {\begin{array}{cccccccccccccccccccc}1&1&1\\0&{\sin \theta }&0\\0&0&{\cos \theta }\end{array}} \right|$
and $\left. {{R_3} \to {R_3} - {R_1}} \right]$
On expanding along third row, we get the value of the determinant
$= \cos \theta \cdot \sin \theta = \frac{1}{2}\sin 2\theta = \frac{1}{2}$
[when $\theta$ is ${45^\circ }$ which gives maximum value]
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Determinants. Curated by Sachin Sharma. Free for all students.