If A=35^{circ}, B=15^{circ} and C=40^{circ},t | vedclass

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(B) Given that,$A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$.Since $A+B+C = 35^{\circ} + 15^{\circ} + 40^{\circ} = 90^{\circ}$,we have $	an(A+B+C)

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$1$

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(B) Given that,$A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$.Since $A+B+C = 35^{\circ} + 15^{\circ} + 40^{\circ} = 90^{\circ}$,we have $	an(A+B+C) = 	an(90^{\circ})$,which is undefined.The formula for $	an(A+B+C)$ is given by:$	an(A+B+C) = \frac{	an A + 	an B + 	an C - 	an A 	an B 	an C}{1 - (	an A 	an B + 	an B 	an C + 	an C 	an A)}$For this to be undefined,the denominator must be zero:$1 - (	an A 	an B + 	an B 	an C + 	an C 	an A) = 0$Therefore,$	an A 	an B + 	an B 	an C + 	an C 	an A = 1$.

If $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$,then $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ is equal to

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