A) \[\Delta \]
B) \[2\Delta \]
C) \[3\Delta \]
D) \[4\Delta \]
Correct Answer: D
Solution :
\[{{a}^{2}}\sin 2C+{{c}^{2}}\sin 2A\]\[={{a}^{2}}(2\sin C\cos C)+{{c}^{2}}(2\sin A\cos A)\] \[=2{{a}^{2}}\left( \frac{2\Delta }{ab}\cos C \right)+2{{c}^{2}}\left( \frac{2\Delta }{bc}\cos A \right)\] \[(\because \,\Delta =\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A,\,\,\,\therefore \,\sin C=\frac{2\Delta }{ab},\,\sin A=\frac{2\Delta }{bc})\] = \[4\Delta \left\{ \frac{a\cos C+c\cos A}{b} \right\}=4\Delta \left( \frac{b}{b} \right)=4\Delta \].
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