If `A , Ba n dC` are the angels of a triangle, show that `|-1+cos B cos C+cos B cos B cos C+cos A-1+cos A cos A-1+cos B-1+cos A-1|=0`

Explanation

`Delta= |{:(-1+cos B,,cos C+cos B,,cos B),(cos C+cos A,,-1+cos A,,cos A),(-1+cos B,,-1+cos A,,-1):}|`
Applying `C_(1) to C_(1) -C_(3):C_(2)to C_(2)-C_(3)`
`|{:(-1,,cos C,,cos B),(cos C,,-1,,cos A),(cos B,,cos A,,-1):}|" "underset("A,B and C respectvely")underset("opposite to angles")("where a, b and c are sides")`
Applying `C_(1) to C_(1)+bC_(2)+cC_(3)`
`=(1)/(a) |{:(-a+b cos C+c cos B,,cos C,,cos B),(a cos C-b + cos A,,-1,,cos A),(a cos B +b cos A-c,,cos A,,-1):}|`
`=(1)/(a) |{:(0,,cos C,,cos B),(0,,-1,,cos A),(0,,cos A,,-1):}|=0`
(using projection rule in trianble)