A) 13
B) 5
C) 7
D) 9
Correct Answer: A
Solution :
[a] \[\frac{\cot A+\cot B}{\cot C}=\frac{\sin (A+B)}{\operatorname{sinAsinB}}.\frac{\sin C}{\cos C}\] \[=\frac{{{\sin }^{2}}C}{\sin A\sin B\cos C}=\frac{{{c}^{2}}}{ab}.\frac{2ab}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}\] \[=\frac{2{{c}^{2}}}{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}=\frac{2{{c}^{2}}}{\frac{17{{c}^{2}}}{9}-{{c}^{2}}}=\frac{9}{4}=\frac{m}{n}\] \[\Rightarrow (m+n)=9+4=13\]
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