A) \[\frac{ab}{2}\]when\[\theta =\frac{\pi }{4}\]
B) \[\frac{3ab}{4}\]when\[\theta =\frac{\pi }{4}\]
C) \[\frac{ab}{2}\]when\[\theta =-\frac{\pi }{2}\]
D) \[{{a}^{2}}{{b}^{2}}\]
Correct Answer: A
Solution :
Area of\[\Delta =\frac{1}{2}\left| \left| \begin{matrix} 0 & 0 & 1 \\ a\cos \theta & b\sin \theta & 1 \\ a\cos \theta & -b\sin \theta & 1 \\ \end{matrix} \right| \right|\] \[\Rightarrow \] \[\Delta =\frac{1}{2}|[1(-ab\sin \theta \cos \theta \]\[-ab\sin \theta \cos \theta )]|\] \[=\frac{ab\sin 2\theta }{2}\] Since, maximum value of\[\sin 2\theta \]is 1, when\[\theta =\frac{\pi }{4}\] \[\therefore \] \[{{\Delta }_{\max }}=\frac{ab}{2}\]
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