In a triangle ABC,if a-2b+c=0,then cot left(f | vedclass

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(C) Given $a - 2b + c = 0$,we have $a + c = 2b$.Using the formula $\cot \left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \left(\

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$3$

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(C) Given $a - 2b + c = 0$,we have $a + c = 2b$.Using the formula $\cot \left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \left(\frac{C}{2}\right) = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$,where $s = \frac{a+b+c}{2}$.Then $\cot \left(\frac{A}{2}\right) \cdot \cot \left(\frac{C}{2}\right) = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \frac{s}{s-b}$.Substitute $s = \frac{a+b+c}{2}$:$\frac{s}{s-b} = \frac{\frac{a+b+c}{2}}{\frac{a+b+c}{2} - b} = \frac{a+b+c}{a+b+c-2b} = \frac{a+b+c}{a-b+c}$.Since $a+c = 2b$,we substitute this into the expression:$\frac{(a+c)+b}{(a+c)-b} = \frac{2b+b}{2b-b} = \frac{3b}{b} = 3$.

In a triangle $ABC$,if $a-2b+c=0$,then $\cot \left(\frac{A}{2}\right) \cdot \cot \left(\frac{C}{2}\right) = $

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