A) 0
B) \[{{a}^{2}}\]
C) \[{{b}^{2}}-{{c}^{2}}\]
D) \[{{a}^{2}}-{{b}^{2}}\]
Correct Answer: C
Solution :
\[a(b\cos C-c\cos B)\] \[=a\left[ b\left( \frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} \right)-c\left( \frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac} \right) \right]\] \[=a\left[ \frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2a}-\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2a} \right]\] \[=\left( \frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2} \right)-\left( \frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2} \right)\] \[=\frac{{{a}^{2}}}{2}+\frac{{{b}^{2}}}{2}-\frac{{{c}^{2}}}{2}-\frac{{{c}^{2}}}{2}-\frac{{{a}^{2}}}{2}+\frac{{{b}^{2}}}{2}\] \[=\frac{2{{b}^{2}}}{2}-\frac{2{{c}^{2}}}{2}\] \[={{b}^{2}}-{{c}^{2}}\]
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