asinA+bcosA=c
Then prove that acosA-bsinA=+-√a^2+b^2-c^2
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(asinA+bcosA)^2=c^2
a^2sin^2A+b^2cos^2A+2absinAcosA=c^2
a^2 (1_cos^2A)+b^2 (1-sin^2A)+2absinAcosA=c^2
a^2-a^2cos^2A+b^2-b^2sin^2A+2absinAcosA=c^2
a^2+b^2-c^2=a^2cos^2A+b^2sin^2A-2absinAcosA
a^2+b^2-c^2=(acosA_bsinA)^2
+-root of a^2+b^2-c^2=(acosA^bsinA)
a^2sin^2A+b^2cos^2A+2absinAcosA=c^2
a^2 (1_cos^2A)+b^2 (1-sin^2A)+2absinAcosA=c^2
a^2-a^2cos^2A+b^2-b^2sin^2A+2absinAcosA=c^2
a^2+b^2-c^2=a^2cos^2A+b^2sin^2A-2absinAcosA
a^2+b^2-c^2=(acosA_bsinA)^2
+-root of a^2+b^2-c^2=(acosA^bsinA)
llado3:
its right statement i got it answer
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2
Answer:
Step-by-step explanation:
(asinA+bcosA)^2=c^2
a^2sin^2A+b^2cos^2A+2absinAcosA=c^2
a^2 (1_cos^2A)+b^2 (1-sin^2A)+2absinAcosA=c^2
a^2-a^2cos^2A+b^2-b^2sin^2A+2absinAcosA=c^2
a^2+b^2-c^2=a^2cos^2A+b^2sin^2A-2absinAcosA
a^2+b^2-c^2=(acosA_bsinA)^2
+-root of a^2+b^2-c^2=(acosA^bsinA)
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