If A - B - C - then begin vmatrix sin A-B-C -sin B -cos C l l-sin B - 0 - tan Allcos A-B -tna A - 0lendvmatrix is equal toA- 1B- 0C- -1D- 2

The correct option is B 0Δ = sin π amp; sin B amp; cos C sin B amp; 0 amp; tan A cos (π C) amp; tna A amp; 0 = 0 amp; sin B amp; ...

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Special Determinants

If A + B + C ...

Question

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

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∣ ∣ ∣ ∣ \begin{matrix} s i n 2 A & s i n C & s i n B s i n C & s i n 2 B & s i n A s i n B & s i n A & s i n 2 C \end{matrix} ∣ ∣ ∣ ∣

A + B + C = π

∣ ∣ ∣ ∣ ∣ \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 a n A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

s i n (B + C) = \frac{\sqrt{3}}{2}

s i n (A + C) = s i n B and

s i n A = \frac{1}{2} .

sin (A + B - C)

A + B + C = π

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

A + B + C = π

sin A + sin B - sin C = 2 sin \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2}

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