prove that using determinants
|sin(A+B+C) sin B cos C |
|-sinb 0 tan A | = 0,if A+B+c=pi?
|cos(A+B) -tan A 0 |

Question

Lovina Kansal · Accepted Answer

Dear student
A+B+C=π⇒A+C=π-B,A+B=π-C and B+C=π-AThus the determinant becomessinπsinπ-BcosC-sinB0tanAcos(π-C)tanπ-A0sinπ=0,sinπ-B=sinB,cosπ-C=-cosC,tanπ-A=-tanAIt is a skew symmetric matrix of the order 3
 Thus, by property of determinats , we get△=0⇒0sinBcosC-sinB0tanA-cosC-tanA0=0
Regards

prove that using determinants |sin(A+B+C) sin B cos C | |-sinb 0 tan A | = 0,if A+B+c=pi? |cos(A+B) -tan A 0 |

prove that using determinants
|sin(A+B+C) sin B cos C |
|-sinb 0 tan A | = 0,if A+B+c=pi?
|cos(A+B) -tan A 0 |