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(D) We know that in $	riangle ABC$,the sum of the three angles is $180^{\circ}$.i.e.,$\angle A + \angle B + \angle C = 180^{\circ}$.Since the triangl

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(D) We know that in $	riangle ABC$,the sum of the three angles is $180^{\circ}$.i.e.,$\angle A + \angle B + \angle C = 180^{\circ}$.Since the triangle is right-angled at $C$,we have $\angle C = 90^{\circ}$ (given).Substituting this value,we get $\angle A + \angle B + 90^{\circ} = 180^{\circ}$.Therefore,$A + B = 180^{\circ} - 90^{\circ} = 90^{\circ}$.Now,we need to find the value of $\cos(A + B)$.Substituting $A + B = 90^{\circ}$,we get $\cos(90^{\circ}) = 0$.

If $\triangle ABC$ is right-angled at $C$,then the value of $\cos(A + B)$ is

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