Question
In a $\triangle\text{ABC},$ Prove that:
$\cos(\text{A+B})= \cos (\pi-\text{C})=0$

Answer

We have: $\text{A + B + C}= \pi$ $\big(\because $ sum of 3 anglesof a triangle is $\pi=180^\circ\big)$
$\Rightarrow \text{A+B}=\pi-\text{C}$
$\Rightarrow \cos(\text{A+B})= \cos (\pi-\text{C})$
$\Rightarrow -\cos\text{C}$ $(\because\cos(\pi-\theta)=-\cos\theta)$
$\Rightarrow \cos(\text{A+B})+\cos\text{C}=0$
$\text{Proved}$

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