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 :
Now \[2\,ca\,\sin \,\left( \frac{A-B+C}{2} \right)\] \[=2\,ca\,\sin \,\left( \frac{\pi }{2}-\frac{B}{2}-\frac{B}{2} \right)\] \[(\because \,\,A+B+C=\pi )\] \[=2\,ca\,\cos B=2ca.\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]
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