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
8
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