If any triangle ABC with usual notations prove `c=a cos B + b cos A`.

Explanation

By cosine rule and with usual notations, we have
`b^(2)=c^(2)+a^(2)-2ca cos B`.
`:.cos B = (c^(2)+a^(2)-b^(2))/(2ca)`
Similarly, `cos A = (b^(2)+c^(2)-a^(2))/(2bc)`
R.H.S. `=a*cos B+b * cos A`
`=a*((c^(2)+a^(2)-b^(2))/(2ca))+b*((b^(2)+c^(2)-a^(2))/(2bc))`
`=(c^(2)+a^(2)-b^(2))/(2c)+(b^(2)+c^(2)-a^(2))/(2c)`
`=(c^(2)+a^(2)-b^(2)+b^(2)+c^(2)-a^(2))/(2c)`
`=(2c^(2))/(2c)`
`=c=` L.H.S.
`:.c=a*cos B+b*cos A " " ` Hence Proved.