Solve qn no 13:


If   A + B + C   = π ,   then   find   the   value   of                         sin   A + B + C   sin   B cost   C - sin   A 0 tan   A cos   A + B - tan   A 0 Using   propertles   of   determinant ,   prove   that                                 b + c a - b a c + a b - c b a + b c - a c = 3 abc - a 3 - b 3 - c 3


Dear Student,
Please find below the solution to the asked query:

A+B+C=πA+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=00sinBcosC-sinB0tanA-cosC-tanA0=0

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