If △ ABC is such that ∠A=90^ , ∠B ∠C, then b^2+c^2b^2-c^2 sin ⁡ (B-C ) is equal to

Question

Accepted Answer

Correct answer: C. (c) In a given triangle ABC, ∠A=90^ , ∠B ∠C ∵ b^2+c^2b^2-c^2 sin ⁡ (B-C ) = k^2 sin^2 ⁡B+k^2 sin^2 ⁡Ck^2 sin^2 ⁡B-k^2 sin^2 ⁡C sin ⁡ (B-C ) ( ∵ a sin ⁡A= b sin ⁡B= c sin ⁡C=k) = sin^2 ⁡B+ sin^2 ⁡C sin^2 ⁡B- sin^2 ⁡C sin ⁡ (B-C ) = sin^2 ⁡B+ sin^2 ⁡(90^ -B) sin ⁡ (B+C ) sin ⁡ (B-C ) sin ⁡ (B-C ) ( ∵B+C=90^ ) = sin^2 ⁡B+ cos^2 ⁡B sin ⁡90^ =1.

$If △ ABC is such that ∠ A = 90^{\circ}, ∠ B \neq = ∠ C, then \frac{b ^{2} + c ^{2}}{b ^{2} - c ^{2}} sin (B - C) is equal to$

Solution

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