is asinA+bcosA=c then prove that bsinA-acosA=+-√a^2+b^2+c^2.
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heya folk !!!
asinA+bcosA=c. (given)------1)
so,,, adding both equation and squaring .
then we get....
(asinA+bcosA)^2+( bsinA-acosA)^2
=)a^2sinA+b^2cos^2+2aSinA*bcosB+b^2sinA+a^2cos^2A-2sinA*cosA
=)a^2sin^2A+a^2cosA+b^2sin^2B+b^2cos^2A
=)a^2(1)+b^2(1)
=)a^2+b^2------------2)
now,,,,, adding again both equation and squaring.
we get ,
(asinA+bcosA)^2+(bsinA-acosA)^2=a^2+b^2. from 2)
=)c^2+(bsinA+acosA)^2=a^2+b^2
=)bsinA+acosA=+_√a^2+b^2-c^2 ,,prooved here .
hope it help you ☺
@rajukumar☺
asinA+bcosA=c. (given)------1)
so,,, adding both equation and squaring .
then we get....
(asinA+bcosA)^2+( bsinA-acosA)^2
=)a^2sinA+b^2cos^2+2aSinA*bcosB+b^2sinA+a^2cos^2A-2sinA*cosA
=)a^2sin^2A+a^2cosA+b^2sin^2B+b^2cos^2A
=)a^2(1)+b^2(1)
=)a^2+b^2------------2)
now,,,,, adding again both equation and squaring.
we get ,
(asinA+bcosA)^2+(bsinA-acosA)^2=a^2+b^2. from 2)
=)c^2+(bsinA+acosA)^2=a^2+b^2
=)bsinA+acosA=+_√a^2+b^2-c^2 ,,prooved here .
hope it help you ☺
@rajukumar☺
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