In triangle ABC right angle at C,find the value of cos(A+B)
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Here in a triangle, given, ✓C=90° angleA+angleB+angleclC=180° . Then, ✓A+✓B=180°-C. => Cos(A+B)= Cos90° => Cos (A+B)=0.
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since ABC is right angled and angle C is 90°therefore,A+B=180° – CA+B=180°-90°A+B= 90°Therefore,cos (A+B)=cos90°=0