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We know that, in ∆ABC,

∠A+∠B+∠C = 180°

But the triangle is right angled at C

ie., ∠C = 90°

Therefore, ∠A+∠B+ 90° = 180°

⇒ ∠A + ∠B = 90°

Therefore, cos(A + B) = cos 90º = 0

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