Question

asinA+bcosA=c
Then prove that acosA-bsinA=+-√a^2+b^2-c^2

Answer 1

(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)

Answer 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)

Read more on Brainly.in - https://brainly.in/question/2074108#readmore