If acosx-bsinx=c,show thatasinx+bcosx=whole root a^2+b^2-c^2

It is given that acosx-bsinx=cNow, squaring both sides, we get(acosx-bsinx)^2=c^2a^2cos^2x+b^2sin^2x-2abcosxsinx=c^2a^2 (1-sin^2x)+b^2 (1-cos^2x)-2abcosxsinx=c^2a^2-a^2sin^2x+b^2-b^2cos^2x-2abcosxsinx =c^2-a^2sin^2x-b^2cos^2x-2abcosxsinx =c^2-a^2-b^2-(asinx+bcosx)^2=c^2-a^2-b^2(asinx+bcosx)^2=a^2+b^2-c^2asinx+bcosx=under root of a^2+b^2-c^2