In any - ABC- prove thatsin B - C -sin B - C - b 2-c2- a 2

By the sine rule, we have a/sin A + b/sin B + c/sin C = k (say) ⇒ a = k sin A, b = k sin B, c = k sin C. ∴ RHS = (b^2 c^2)/a^2 = k^2 sin^2 B k^2 sin^2 C/sin ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

In any Δ ABC,...

Question

In any $Δ A B C,$ prove that

$\frac{s i n (B - C)}{s i n (B + C)} = \frac{(b^{2} - c^{2})}{a^{2}}$

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Solution

By the sine rule, we have

$\frac{a}{s i n A} + \frac{b}{s i n B} + \frac{c}{s i n C} = k (s a y)$

$\Rightarrow a = k s i n A, b = k s i n B, c = k s i n C .$

$∴ R H S = \frac{(b^{2} - c^{2})}{a^{2}} = \frac{k^{2} s i n^{2} B - k^{2} s i n^{2} C}{s i n^{2} (B + C)}$

= $\frac{(s i n^{2} B - s i n^{2} C)}{s i n^{2} A} = \frac{s i n (B + C) . s i n (B - C)}{s i n^{2} (B + C)}$

$[\begin{matrix} ∵ (A + B + C) = π \Rightarrow A = π - (B + C) ∴ s i n A = s i n | π - (B + C) | = s i n (B - C) \end{matrix}]$

= $\frac{s i n (B - C)}{s i n (B + C)} = L H S$

$∴$ RHS = LHS

Hence, $\frac{s i n (B - C)}{s i n (B + C)} = s i n \frac{(A - B)}{2}$

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In any Δ ABC, prove that sin(B−C)sin(B+C)=(b2−c2)a2

In any $Δ A B C,$ prove that

$\frac{s i n (B - C)}{s i n (B + C)} = \frac{(b^{2} - c^{2})}{a^{2}}$