If A + B + C = frac{pi}{2},then the value of | vedclass

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(C) Given that $A + B + C = \frac{\pi}{2} = 90^\circ$.Therefore,$A + B = 90^\circ - C$.Taking the tangent on both sides:$	an(A + B) = 	an(90^\circ -

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$1$

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(C) Given that $A + B + C = \frac{\pi}{2} = 90^\circ$.Therefore,$A + B = 90^\circ - C$.Taking the tangent on both sides:$	an(A + B) = 	an(90^\circ - C)$Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(90^\circ - C) = \cot C = \frac{1}{	an C}$,we get:$\frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{1}{	an C}$Cross-multiplying:$	an C(	an A + 	an B) = 1 - 	an A 	an B$Expanding the left side:$	an C 	an A + 	an C 	an B = 1 - 	an A 	an B$Rearranging the terms:$	an A 	an B + 	an B 	an C + 	an C 	an A = 1$.

If $A + B + C = \frac{\pi}{2}$,then the value of $\tan A \tan B + \tan B \tan C + \tan C \tan A$ is

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