Question

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

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $0$

Since ABC is a right angled triangle right angled at C
Hence
$A+B={180}^0-C={90}^0$
Hence
$cos(A+B)$
$=cos\left({90}^0\right)$
$=0$