In ΔABC, prove that `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")` = 0

In ∆ABC by sine rule, we have `(sin"A")/"a" = (sin"B")/"b" = (sin"C")/"c"` = k &there4; sin A = ka, sin B = kb, sin C = kc Consider a2sin(B − C) = a2(sin B cos C − cos B sin C) = a2(kb cos C − kc cos B) = ka(ab cos C − ac cos B) = `"ak"["ab"(("a"^2 + "b"^2 - "c"^2)/(2"ab")) - "ac"(("a"^2 + "c"^2 - "b"^2)/(2"ac"))]` .......[By consine rule] = `"ak"[("a"^2 + "b"^2 + "c"^2)/2 - (("a"^2 + "c"^2 - "b"^2)/2)]` = `"k"/2 ("a")("a"^2 + "b"^2 - "c"^2 - "a"^2 - "c"^2 + "b"^2)` = `"k"/2 "a"(2"b"^2 - 2"c"^2)` = ka(b2 − c2) Similarly, we can prove that b2sin(C − A) = kb(c2 − a2) and c2sin(A − B) = kc(a2 − b2) &there4; `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")` = `("ka"("b"^2 - "c"^2))/(sin"A") + ("kb"("c"^2 - "a"^2))/(sin"B") + ("kc"("a"^2 - "b"^2))/(sin"C")` = `("ka"("b"^2 - "c"^2))/"ka" + ("kb"("c"^2 - "a"^2))/"kb" + ("kc"("a"^2 - "b"^2))/"kc"` = (b2 − c2 + c2 − a2 + a2 − b2) = 0 &there4; `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")` = 0

In ΔABC, prove that aBCAbCABcABCa2sin(B-C)sinA+b2sin(C-A)sinB+c2sin(A-B)sinC = 0 - Mathematics and Statistics

प्रश्न

In ΔABC, prove that $\frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C}$ = 0

योग

उत्तर

In ∆ABC by sine rule, we have

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ = k

∴ sin A = ka, sin B = kb, sin C = kc

Consider a²sin(B − C) = a²(sin B cos C − cos B sin C)

= a²(kb cos C − kc cos B)

= ka(ab cos C − ac cos B)

= $ak [ab (\frac{a^{2} + b^{2} - c^{2}}{2 ab}) - ac (\frac{a^{2} + c^{2} - b^{2}}{2 ac})]$ .......[By consine rule]

= $ak [\frac{a^{2} + b^{2} + c^{2}}{2} - (\frac{a^{2} + c^{2} - b^{2}}{2})]$

= $\frac{k}{2} (a) (a^{2} + b^{2} - c^{2} - a^{2} - c^{2} + b^{2})$

= $\frac{k}{2} a (2 b^{2} - 2 c^{2})$

= ka(b² − c²)

Similarly, we can prove that

b²sin(C − A) = kb(c² − a²) and c²sin(A − B)

= kc(a² − b²)

∴ $\frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C}$

= $\frac{ka (b^{2} - c^{2})}{\sin A} + \frac{kb (c^{2} - a^{2})}{\sin B} + \frac{kc (a^{2} - b^{2})}{\sin C}$

= $\frac{ka (b^{2} - c^{2})}{ka} + \frac{kb (c^{2} - a^{2})}{kb} + \frac{kc (a^{2} - b^{2})}{kc}$

= (b² − c² + c² − a² + a² − b²)

= 0

∴ $\frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C}$ = 0

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