If A , B , C are the angles of a triangle and 1 1 1 1 + sin ⁡ A 1 + sin ⁡ B 1 + sin ⁡ C sin ⁡ A + sin 2 ⁡ A sin ⁡ B + sin 2 ⁡ B sin ⁡ C + sin 2 ⁡ C = 0, then triangle A B C is

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Correct answer: B. isosceles

R 2 → R 2 - R 1 1 1 1 sin ⁡ A sin ⁡ B sin ⁡ C sin ⁡ A + sin 2 ⁡ A sin ⁡ B + sin 2 ⁡ B sin ⁡ C + sin 2 ⁡ C R 3 → R 3 - R 2 = 1 1 1 sin ⁡ A sin ⁡ B sin ⁡ C sin 2 ⁡ A sin 2 ⁡ B sin 2 ⁡ C = sin ⁡ A - sin ⁡ B × sin ⁡ B - sin ⁡ C × sin ⁡ C - sin ⁡ A ⇒ Either A = B or B = C or C = A So Δ is isosceles

If A , B , C are the angles of a triangle and 1 1 1 1 + sin ⁡ A 1 + sin ⁡ B 1 + sin ⁡ C sin ⁡ A + sin 2 ⁡ A sin ⁡ B + sin 2 ⁡ B sin ⁡ C + sin 2 ⁡ C = 0, then triangle A B C is

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