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If $f(\theta ) =\left| {\begin{array}{*{20}{c}}
1&{\cos {\mkern 1mu} \theta }&1\\
{ - \sin {\mkern 1mu} \theta }&1&{ - \cos {\mkern 1mu} \theta }\\
{ - 1}&{\sin {\mkern 1mu} \theta }&1
\end{array}} \right|$ and $A$ and $B$ are respectively the maximum and the minimum values of $f(\theta )$, then $(A , B)$ is equal to
$(3, - 1)$
$( 4,2-\sqrt 2 )$
$(2 + \sqrt 2 ,2 - \sqrt 2 )$
$(2 + \sqrt 2 , - 1)$
Solution
Let $f\left( \theta \right) = \begin{array}{*{20}{c}}
1&{\cos \theta }&1\\
{ – \sin \theta }&1&{ – \cos \theta }\\
{ – 1}&{\sin \theta }&1
\end{array}$
$ = \left( {1 + \sin \theta \cos \theta } \right) – \cos \theta \left( {\sin \theta – \cos \theta } \right) + 1\left( { – {{\sin }^2}\theta + 1} \right)$
$ = 1 + \sin \theta \cos \theta + \sin \theta \cos \theta + {\cos ^2}\theta – {\sin ^2}\theta + 1$
$ = 2 + 2\sin \theta \cos \theta + \cos 2\theta $
$ = 2 + \sin 2\theta + \cos 2\theta \,\,\,\,\,\,\,\,\,……\left( 1 \right)$
Now, maximum value of $(1)$
is $2 + \sqrt {{1^2} + {1^2}} = 2 + \sqrt 2 $
and minimum value of $(1)$ is
$2 – \sqrt {{1^2} + {1^2}} = 2 – \sqrt 2 $.
