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ABC-triangle-s-A-B-C-90-prove-that-sin-2-A-sin-2-B-sin-2-C-0-

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Question Number 131785 by faysal last updated on 08/Feb/21
ABC triangle′s A+B=C=90° prove that sin^2 A−sin^2 B+sin^2 C=0
ABCtriangle′sA+B=C=90°provethatsin2A−sin2B+sin2C=0
Answered by Dwaipayan Shikari last updated on 08/Feb/21
(a/(sinA))=(b/(sinB))=(c/(sinC))=φ sin^2 A−sin^2 B+sin^2 C=((a^2 +c^2 −b^2 )/φ^2 )=Φ A+B=C=(π/2) ⇒a^2 +c^2 =b^2 Φ=((b^2 −b^2 )/φ^2 )=0
asinA=bsinB=csinC=ϕsin2A−sin2B+sin2C=a2+c2−b2ϕ2=ΦA+B=C=π2⇒a2+c2=b2Φ=b2−b2ϕ2=0

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