Let A, B and C be the angles of a plane trian | vedclass

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

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$7/9$

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(A) In a triangle $ABC$,the sum of angles is $A + B + C = \pi$.Dividing by $2$,we get $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.Taking the tangent on both sides,we have $	an \left( \frac{A+B}{2} ight) = 	an \left( \frac{\pi}{2} - \frac{C}{2} ight) = \cot \frac{C}{2}$.Using the formula $	an(x+y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we get $\frac{	an \frac{A}{2} + 	an \frac{B}{2}}{1 - 	an \frac{A}{2} 	an \frac{B}{2}} = \cot \frac{C}{2}$.Substituting the given values $	an \frac{A}{2} = \frac{1}{3}$ and $	an \frac{B}{2} = \frac{2}{3}$:$\cot \frac{C}{2} = \frac{\frac{1}{3} + \frac{2}{3}}{1 - (\frac{1}{3})(\frac{2}{3})} = \frac{1}{1 - \frac{2}{9}} = \frac{1}{\frac{7}{9}} = \frac{9}{7}$.Since $\cot \frac{C}{2} = \frac{9}{7}$,we have $	an \frac{C}{2} = \frac{7}{9}$.

Let $A, B$ and $C$ be the angles of a plane triangle and $\tan \frac{A}{2} = \frac{1}{3}, \tan \frac{B}{2} = \frac{2}{3}$. Then $\tan \frac{C}{2}$ is equal to

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