If A+B+C=π, then ∣∣ ∣ ∣∣sin(A+B+C)sinBc

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If

A + B + C = π

, then

∣ ∣∣∣ ∣ \begin{matrix} s i n (A + B + C) & s i n B & c o s C - s i n B & 0 & t a n A c o s (A + B) & - t n a A & 0 \end{matrix} ∣ ∣∣∣ ∣

is equal to

Options:

A . 1

B . 0

C . -1

D . 2

Answer: Option B
:
B

Δ = ∣ ∣∣ ∣ \begin{matrix} s i n π & s i n B & c o s C - s i n B & 0 & t a n A c o s (π - C) & - t n a A & 0 \end{matrix} ∣ ∣∣ ∣ = ∣ ∣∣ ∣ \begin{matrix} 0 & s i n B & c o s C - s i n B & 0 & t a n A - c o s C & - t a n A & 0 \end{matrix} ∣ ∣∣ ∣

=0 (

∵ Δ

is skew symmetric)

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