A) \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}\]
B) \[{{c}^{2}}+{{a}^{2}}-{{b}^{2}}\]
C) \[{{b}^{2}}-{{c}^{2}}-{{a}^{2}}\]
D) \[{{c}^{2}}-{{a}^{2}}-{{b}^{2}}\]
Correct Answer: B
Solution :
We know that \[A+B+C=\pi \] \[\Rightarrow \] \[A+C=\pi -B\] \[\Rightarrow \] \[\frac{A-B+C}{2}=\frac{\pi }{2}-B\] \[\therefore 2\operatorname{c}a\sin \left( \frac{A-B+C}{2} \right)=2ca\sin \left( \frac{\pi }{2}-B \right)\] \[=2ac\cos B=2ac\left( \frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2\,ac} \right)\] \[={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]
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