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(B) Since $A, B, C$ are angles of a triangle,$A+B+C = \pi$.Therefore,$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A+

Vedclass · Accepted Answer

$\frac{7}{9}$

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(B) Since $A, B, C$ are angles of a triangle,$A+B+C = \pi$.Therefore,$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.Taking tangent on both sides,$	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}$,we have:$\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}$:$\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}$.Thus,$\cot \frac{C}{2} = \frac{9}{7}$,which implies $	an \frac{C}{2} = \frac{7}{9}$.

If $A, B, C$ are the angles of a triangle with $\tan \frac{A}{2}=\frac{1}{3}$ and $\tan \frac{B}{2}=\frac{2}{3}$,then the value of $\tan \frac{C}{2}$ is:

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