A) 0
B) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
C) \[2\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]
D) \[\frac{1}{2abc}\]
Correct Answer: A
Solution :
\[({{b}^{2}}-{{c}^{2}})\cot A=({{b}^{2}}-{{c}^{2}})\frac{\cos A}{\sin A}=\frac{({{b}^{2}}-{{c}^{2}})({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2bc.ka}\] Hence L.H.S. \[=\frac{1}{2k\,\,abc}\] \[[({{b}^{4}}-{{c}^{4}})+({{c}^{4}}-{{a}^{4}})+({{a}^{4}}-{{b}^{4}})-\{{{a}^{2}}({{b}^{2}}-{{c}^{2}})\] \[+{{b}^{2}}({{c}^{2}}-{{a}^{2}})+{{c}^{2}}({{a}^{2}}-{{b}^{2}})\}]=0\].
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