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Q.
The maximum value of $\Delta=\left|\begin{matrix}1&1&1\\ 1&1+sin\,\theta&1\\ 1+cos\,\theta&1&1\end{matrix}\right|$ is ($\theta$ is real number)
Determinants
Solution:
$\Delta=\left|\begin{matrix}1&1&1\\ 1&1+sin\,\theta&1\\ 1+cos\,\theta&1&1\end{matrix}\right|$
Applying $R_{1}\rightarrow R_{1}-R_{3} and R_{2}\rightarrow R_{2}-R_{3}$, we get
$\Delta=\left|\begin{matrix}-cos\,\theta&0&0\\ -cos\,\theta&sin\,\theta&0\\ 1+cos\,\theta&1&1\end{matrix}\right|$
Expanding along $C_{3}$, we get
$\Delta=1\left(-cos\,\theta sin\,\theta-0\right)=-cos\,\theta sin\,\theta=\frac{-1}{2}sin\,2\theta$
We know, $- 1 \le sin 2\theta \le 1$
$\Rightarrow \quad-\frac{1}{2}\le\frac{1}{2} sin\,2\theta \le\frac{1}{2}\quad\Rightarrow \quad\frac{1}{2}\ge-\frac{1}{2} sin\,2\theta \ge\frac{-1}{2}$
i.e., $-\frac{1}{2} \le-\frac{1}{2} sin\,2\theta \le \frac{1}{2}\quad\Rightarrow \quad\frac{-1}{2}\le\Delta \le\frac{1}{2}$
Hence, maximum value of $\Delta$ is $\frac{1}{2}$.