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 use the identity for $

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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 use the identity for $	an(A+B+C)$:$	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)}$.As $A+B+C = 90^{\circ}$,$	an(90^{\circ})$ is undefined,which implies 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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