In a triangle ABC,tan frac{A}{2} tan frac{B}{ | vedclass

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(B) In a triangle $ABC$,$A+B+C = \pi$. Dividing by $2$,we get $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A}{2} + \

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(B) In a triangle $ABC$,$A+B+C = \pi$. Dividing by $2$,we get $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}$. Taking tangent on both sides: $	an(\frac{A}{2} + \frac{B}{2}) = 	an(\frac{\pi}{2} - \frac{C}{2}) = \cot \frac{C}{2}$. Using the identity $	an(x+y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we have $\frac{	an \frac{A}{2} + 	an \frac{B}{2}}{1 - 	an \frac{A}{2} 	an \frac{B}{2}} = \frac{1}{	an \frac{C}{2}}$. Cross-multiplying gives: $	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$.

In a triangle $ABC$,$\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} =$

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