If A + B + C = pi (A, B, C 0) and the angle | vedclass

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(B) Given $A + B + C = \pi$,so $A + B = \pi - C$.Taking tangent on both sides: $	an(A + B) = 	an(\pi - C)$.Using the formula $	an(A + B) = \frac{

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$	an A 	an B < 1$

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(B) Given $A + B + C = \pi$,so $A + B = \pi - C$.Taking tangent on both sides: $	an(A + B) = 	an(\pi - C)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$\frac{	an A + 	an B}{1 - 	an A 	an B} = -	an C$.Since $C$ is an obtuse angle $(90^{\circ}  0$.Therefore,$\frac{	an A + 	an B}{1 - 	an A 	an B} > 0$.Since $A, B, C$ are angles of a triangle and $C$ is obtuse,$A$ and $B$ must be acute,so $	an A > 0$ and $	an B > 0$,implying $	an A + 	an B > 0$.For the fraction to be positive,the denominator must be positive: $1 - 	an A 	an B > 0$.Thus,$	an A 	an B < 1$.

If $A + B + C = \pi$ $(A, B, C > 0)$ and the angle $C$ is obtuse,then:

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