If in a triangle ABC the sides AB and AC are | vedclass

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(D) Given that sides $AB$ and $AC$ are perpendicular,$\angle A = 90^{\circ}$ or $\frac{\pi}{2}$.In any triangle $ABC$,the sum of angles is $A + B + C

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$	an B \cdot 	an C = 1$

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(D) Given that sides $AB$ and $AC$ are perpendicular,$\angle A = 90^{\circ}$ or $\frac{\pi}{2}$.In any triangle $ABC$,the sum of angles is $A + B + C = 180^{\circ}$ or $\pi$.Since $A = 90^{\circ}$,we have $B + C = 90^{\circ}$,which implies $C = 90^{\circ} - B$.Taking the tangent on both sides,$	an C = 	an(90^{\circ} - B)$.Using the trigonometric identity $	an(90^{\circ} - 	heta) = \cot 	heta$,we get $	an C = \cot B$.Since $\cot B = \frac{1}{	an B}$,we have $	an C = \frac{1}{	an B}$.Therefore,$	an B \cdot 	an C = 1$.

If in a triangle $ABC$ the sides $AB$ and $AC$ are perpendicular,then the true equation is

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