In a triangle ABC,if cot frac{A}{2} cot frac{ | vedclass

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(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$.Multiplying these,we

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$(1, \infty)$

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(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$.Multiplying these,we get $\cot \frac{A}{2} \cot \frac{B}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-b)}{(s-a)(s-c)}} = \sqrt{\frac{s^2}{(s-c)^2}} = \frac{s}{s-c}$.Given $\frac{s}{s-c} = K$,we can write $K = \frac{s-c+c}{s-c} = 1 + \frac{c}{s-c}$.Since $s = \frac{a+b+c}{2}$,we have $s-c = \frac{a+b-c}{2}$.Thus,$K = 1 + \frac{c}{\frac{a+b-c}{2}} = 1 + \frac{2c}{a+b-c}$.By the triangle inequality,$a+b > c$,so $a+b-c > 0$,which implies $K > 1$.As $C 	o 180^{\circ}$,$a+b-c 	o 0^{+}$,so $K 	o \infty$.Therefore,the range of $K$ is $(1, \infty)$.

In a triangle $ABC$,if $\cot \frac{A}{2} \cot \frac{B}{2} = K$,then all the possible values of $K$ lie in

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