Q. 13

# If A + B + C =π t

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: --

= (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

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