Question

In a \( \triangle ABC \), prove that :

\[ \cos(A+B)+\cos C=0 \]

Solution

In a triangle,

\[ A+B+C=\pi \]

\[ A+B=\pi-C \]

Therefore,

\[ \begin{aligned} \cos(A+B)+\cos C &= \cos(\pi-C)+\cos C \\[8pt] &= -\cos C+\cos C \\[8pt] &= 0 \end{aligned} \]

Hence Proved.