Question

$\Delta =\left|\begin{array}{ccc}1& 3cos\theta & 1\\ sin\theta & 1& 3cos\theta \\ 1& sin\theta & 1\end{array}\right|=R_1\to R_1-R_3\Delta =\left|\begin{array}{ccc}0& 3cos\theta -sin\theta & 0\\ sin\theta & 1& 3cos\theta \\ 1& sin\theta & 1\end{array}\right|$

Answer 1

$=(sin\theta -3cos\theta )(sin\theta -3cos\theta )$
$\Delta =(sin\theta -3cos\theta {)}^2$
$\frac{d\Delta }{d\theta }=2(sin\theta -3cos\theta )(cos\theta -3sin\theta )=0$
$\therefore Eithersin\theta -3cos\theta =0\Rightarrow tan\theta =3$
or $cos\theta +3sin\theta =0\Rightarrow tan\theta =-\frac{1}{3}$
Now, $sin\theta =\pm \frac{tan\theta }{\sqrt{1+ta{n}^2\theta }}$
$si{n}^2\theta =\frac{ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{a}{10}and\frac{\frac{1}{a}}{1+\frac{1}{9}}=\frac{1}{10}$
at $tan\theta =3$ at $tan\theta =-\frac{1}{3}$
$co{s}^2\theta =\frac{1}{1+ta{n}^2\theta }\Rightarrow \frac{1}{10}and\frac{1}{1+\frac{1}{9}}=\frac{9}{10}$
at $tan\theta =3$ at $tan\theta =-\frac{1}{3}$
$\Delta =(sin\theta -3cos\theta {)}^2=si{n}^2\theta +9co{s}^2\theta -6sin\theta cos\theta$
$=\frac{9}{10}+\frac{9}{10}-6\times \frac{3}{10}\times \frac{1}{10}$
$=\frac{9}{10}+\frac{9}{10}-\frac{18}{10}=0$
$=\frac{1}{10}+\frac{8}{10}+6\times \frac{1}{\sqrt{10}}\times \frac{3}{\sqrt{10}}=\frac{1+8+18}{10}=\frac{100}{10}=10$
$\therefore Maxvalueof\Delta =10$