If A + B + C = 180^o,then the value of cot fr | vedclass

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(C) Given $A + B + C = 180^o$,we have $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.Taking $\cot$ on both sides: $\cot(\frac{A}{2} + \frac{B}{2}) =

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$\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$

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(C) Given $A + B + C = 180^o$,we have $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.Taking $\cot$ on both sides: $\cot(\frac{A}{2} + \frac{B}{2}) = \cot(90^o - \frac{C}{2})$.Using the formula $\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$,we get $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = 	an \frac{C}{2}$.Since $	an \frac{C}{2} = \frac{1}{\cot \frac{C}{2}}$,we have $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = \frac{1}{\cot \frac{C}{2}}$.Cross-multiplying gives $(\cot \frac{A}{2} \cot \frac{B}{2} - 1) \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.Expanding this,we get $\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} - \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.Rearranging the terms,we find $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$.

If $A + B + C = 180^o$,then the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$ is equal to

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