A) 1
B) \[\frac{ab}{4\Delta }\]
C) 0
D) \[\frac{ac}{4\Delta }\]
Correct Answer: C
Solution :
\[\cot B+\cot C-\cot A=\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}-\cot A\] \[=\frac{\sin C\cos B+\cos C\sin B}{\sin B\sin C}-\cot A\]\[=\frac{\sin (B+C)}{\sin B\sin C}-\frac{\cos A}{\sin A}\] \[=\frac{{{\sin }^{2}}A-\sin B\sin C\cos A}{\sin A\sin B\sin C}=\frac{{{a}^{2}}-bc\cos A}{k(abc)}\] Since\[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\] (say) and \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]\[=\frac{{{a}^{2}}-bc\frac{({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2bc}}{(abc)k}\] \[=\frac{({{a}^{2}}-{{a}^{2}})}{abc\,k}=0,\text{ }\left\{ \text{As}\,\,\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2}=\frac{3{{a}^{2}}-{{a}^{2}}}{2}=\frac{2{{a}^{2}}}{2}={{a}^{2}} \right\}.\]
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