For any - ABC- the value of determinant - sin 2 A A 1 sin 2 B B 1 sin 2 C C 1 - is equal to

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Quadratic Formula

For any Δ A...

Question

For any $Δ A B C$ , the value of determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} {sin}^{2} A & cot A & 1 {sin}^{2} B & cot B & 1 {sin}^{2} C & cot C & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

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sin A sin B sin C

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sin A + sin B + sin C

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Solution

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A + B + C = π

f (A, B, C) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {sin}^{2} A & cot A & 1 {sin}^{2} B & cot B & 1 {sin}^{2} C & cot C & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ A B C

s i n A + s i n B + s i n C =

In any $Δ A B C$ , prove that.

$a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = 0$

Δ A B C

\frac{sin A + sin B}{sin C} =

Δ A B C

\sum \frac{\sqrt{sin} A}{\sqrt{sin B} + \sqrt{sin C} - \sqrt{sin A}} \geq 3

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