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

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(B) Given that $A+B+C=180^{\circ}$.We can write $A+B = 180^{\circ}-C$.Dividing both sides by $2$,we get $\frac{A+B}{2} = \frac{180^{\circ}-C}{2} = 90^

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$\cot \frac{C}{2}$

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(B) Given that $A+B+C=180^{\circ}$.We can write $A+B = 180^{\circ}-C$.Dividing both sides by $2$,we get $\frac{A+B}{2} = \frac{180^{\circ}-C}{2} = 90^{\circ}-\frac{C}{2}$.Now,taking the tangent on both sides: $	an \left(\frac{A+B}{2}ight) = 	an \left(90^{\circ}-\frac{C}{2}ight)$.Using the trigonometric identity $	an(90^{\circ}-	heta) = \cot 	heta$,we get $	an \left(\frac{A+B}{2}ight) = \cot \frac{C}{2}$.

If $A+B+C=180^{\circ},$ then $\tan \left(\frac{A+B}{2}\right)=$ ...........

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