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Skew Symmetric Matrix

If A+B+C=π,...

Question

If $A + B + C = π$ , then $∣ ∣ ∣ ∣ ∣ \begin{matrix} sin (A + B + C) & sin B & cos C sin B & 0 & tan A cos (A + B) & - tan A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ =$ _______.

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Solution

Ans. Zero.
Given is skew symmetric determinant of odd order because $cos (A + B) = - cos C$ etc.

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A + B + C = π

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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} ∣ ∣ ∣ ∣ ∣ =

A + B + C = π

∣ ∣ ∣ ∣ ∣ \begin{matrix} tan (A + B + C) & tan B & tan C tan (A + C) & 0 & tan A tan (A + B) & - tan A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

Q. If A + B + C =

π

, then prove that,

s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} = 1 + 4 s i n [\frac{π - A}{4}] s i n [\frac{π - B}{4}] s i n [\frac{π - C}{4}]

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