Question

If $f\left(q\right)=\left|\begin{array}{ccc}1& \cos \theta & 1\\ \sin \theta & 1& \cos \theta \\ 1& \sin \theta & 1\end{array}\right|$ and $A$ and $B$ are respectively the maximum and minimum values of $f\left(\theta \right)$, then $(A,B)$ is equal

Answer 1

Expanding the determinant along row 1,

$f\left(\theta \right)=1(1-\sin \theta \cos \theta )-1(\sin \theta -\cos \theta )+1({\sin }^2\theta -1)$

$=1-\sin \theta \cos \theta -\sin \theta +\cos \theta +{\sin }^2\theta -1$

$=\sin \theta (\sin \theta -\cos \theta )-1(\sin \theta -\cos \theta )$

$=(\sin \theta -1)(\sin \theta -\cos \theta )$

For maximum or minimum, $f^{\prime }\left(\theta \right)=0$

$f^{\prime }\left(\theta \right)=(\sin \theta -1)(-\cos \theta +\sin \theta )+\cos \theta (\sin \theta -\cos \theta )$

$=(\sin \theta -\cos \theta )(\sin \theta +\cos \theta -1)=0$

When $\sin \theta =\cos \theta$ or $\sin \theta +\cos \theta =1$

$\Rightarrow \theta =\frac{\pi }{4}$

$\Rightarrow \theta =0$ or $\frac{\pi }{2}$

At $\theta =0,f\left(\theta \right)=(-1)(-1)=1$

$\theta =\frac{\pi }{4},f\left(\theta \right)=(\frac{1}{\sqrt{2}}-1)(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})=0$

$\theta =\frac{\pi }{2},f\left(\theta \right)=(1-1)(1-0)=0$

$\Rightarrow A=1,B=o$

$\Rightarrow (A,B)$ are not equal.