If A + B + C = 180^o,then sum tan frac{A}{2} | vedclass

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

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

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

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