if a cosß - b sinß = c , prove that : a sinß + b cosß = ± √a² + b² - c²

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`acosbeta-bsinbeta = c`

Take the square value of both sides,

`(acosbeta-bsinbeta)^2 = c^2`

`a^2cos^2beta -2abcosbetasinbeta+b^2sin^2beta = c^2`

Let,

`asinbeta-bcosbeta = d`

`(asinbeta-bcosbeta)^2 = d^2`

`a^2sin^2beta + 2abcosbetasinbeta+b^2cos^2beta = d^2`

Add them together,

`a^2cos^2beta -2abcosbetasinbeta+b^2sin^2beta + a^2sin^2beta + 2abcosbetasinbeta+b^2cos^2beta = c^2+d^2`

This gives,

`a^2cos^2beta + a^2sin^2beta + b^2sin^2beta + b^2cos^2beta = c^2+d^2`

`a^2(cos^2beta + sin^2beta) + b^2(sin^2beta + cos^2beta) = c^2+d^2`

`a^2(1)+b^2(1) = c^2+d^2`

`c^2+d^2 = a^2+b^2`

Therefore,

`d^2 = a^2+b^2-c^2`

`d = +-sqrt(a^2+b^2-c^2)`

Therefore,

`asinbeta-bcosbeta = +-sqrt(a^2+b^2-c^2)`

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