In a triangle `ABC`, if `a,b,c` are the sides opposite to angles `A`, `B`, `C` respectively, then the value of `|{:(bcosC,a,c cosB),(c cosA,b,acosC),(acosB,c,bcosA):}|` is

Options

A. `1`
B. `-1`
C. `0`
D. `acosA+bcosB+c cosC`

Answer

C

Explanation

`(c )` `|{:(bcosC,a,c cosB),(c cosA,b,acosC),(acosB,c,bcosA):}|`
Apply `C_(1)toC_(1)+C_(3)` and using projection rule.
`D=|{:(bcosC,a,c cosB+bcosC),(c cosA,b,acosC+c cosA),(acosB,c,bcosA+acosB):}|`
`=|{:(bcosC,a,a),(c cosA,b,b),(acosB,c,c):}|=0`