If A + B + C =π then find the value of

OR

Using properties of determinant, prove that

_{As A+B+C = π}

_{Then A + B = π – c}

As sin π = 0 and cos (π - θ) = - cos θ

Expanding along C₁ we get,

= sin B (tan A cos C) – cos C (sin B tan A)

= sin B tan A cos C – cos C sin B tan A

= 0

Hence value of .

OR

Applying R₁→ R₁ + R₂ + R₃ in the given determinant, we get

Applying C₁→ C₁ – 2C₃

Expanding along the R₁, we get the determinant as:

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= (a + b + c)[(a + c – 2b)(c – a) – (b – c)(a + b – 2c)]

= (a + b + c)(ac – a² + c² – ac – 2bc + 2ab – ab – b² + 2bc + ca + cb – 2c²)

= (a + b + c)(ab + ac + cb – a² – b² – c²)

= -(a + b + c)(a² + b² + c² – ab – bc - cb)

= - (a³ + b³ + c³ – 3abc)

= 3abc – a³ – b³ – c³