If A + B + C = 180^circ,then sum tan frac{A}{ | vedclass

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(B) Given $A + B + C = 180^\circ$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$.Thus,$\frac{A}{2} = 90^\circ - (\frac{B}{2} + \frac{C}{

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(B) Given $A + B + C = 180^\circ$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$.Thus,$\frac{A}{2} = 90^\circ - (\frac{B}{2} + \frac{C}{2})$.Taking tangent on both sides,$	an \frac{A}{2} = 	an(90^\circ - (\frac{B}{2} + \frac{C}{2})) = \cot(\frac{B}{2} + \frac{C}{2})$.Using the identity $\cot 	heta = \frac{1}{	an 	heta}$,we get $	an \frac{A}{2} = \frac{1}{	an(\frac{B}{2} + \frac{C}{2})}$.Therefore,$	an \frac{A}{2} 	an(\frac{B}{2} + \frac{C}{2}) = 1$.Expanding the tangent sum formula: $	an \frac{A}{2} \left( \frac{	an \frac{B}{2} + 	an \frac{C}{2}}{1 - 	an \frac{B}{2} 	an \frac{C}{2}} ight) = 1$.$	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{A}{2} 	an \frac{C}{2} = 1 - 	an \frac{B}{2} 	an \frac{C}{2}$.Rearranging the terms,we get $	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} = 1$.

If $A + B + C = 180^\circ$,then $\sum \tan \frac{A}{2} \tan \frac{B}{2} = $

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