Answer:
Step-by-step explanation:
(cos A+ cos B)-cos C
[tex]=2cos \frac{A+B}{2}cos \frac{A-B}{2}-cos C~~~...(1)\\A+B+C=180\\A+B=180-C\\\frac{A+B}{2}=90-\frac{C}{2}\\cos \frac{A+B}{2}=cos(90-\frac{C}{2})=sin \frac{C}{2}\\cos C=1-2sin^2\frac{C}{2}\\(1)=2 sin \frac{C}{2} cos \frac{A-B}{2}-1+2sin^2\frac{C}2}\\=2sin\frac{C}{2}[cos \frac{A-B}{2}+sin \frac{C}{2}]-1~~~...(2)\\\\now~again~A+B+C=180\\C=180-(A+B)\\sin\frac{C}{2}=sin(90-\frac{A+B}{2})=cos \frac{A+B}{2}\\(2)=2sin\frac {C}{2}[cos \frac{A-B}{2}+cos \frac{A+B}{2}]-1\\[/tex]
[tex]=-1+2sin\frac{C}{2}*2cos \frac{\frac{A-B}{2} +\frac{A+B}{2} }{2} cos \frac{\frac{A-B}{2} -\frac{A+B}{2} }{2} \\=-1+4sin\frac{C}{2} cos \frac{A}{2} cos\frac{-B}{2} \\=-1+4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}\\(cos(-B)=cos B)[/tex]
