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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, <u>cos(A + B) = cos 90º = 0</u>
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, <u>cos(A + B) = cos 90º = 0</u>