Question

prove that, COS(A+B) +COS C=0​

Answer 1

cos(A+B)+cosC=0

[Therefore,Triangle of ABC is A,B,C

Therefore,A+ B+C=180°

»A+B=180°-C]

Therefore,L. H.S.=cos(A+B)+cosC

=cos(180°-C)+cosC

=-cosC+cosC

=0

=R.H.S.

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