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

Given: ∠C = 90°

By the property of triangle, the sum of the three angles is equal to

∠A+∠B+∠C = 180°

⇒ ∠A+∠B+90° = 180° (∵ ∠C = 90° )

⇒ ∠A+∠B = 90°

Thus, cos(A+B) = cos 90° = 0