If a cos theta +b sin theta =c, then prove that a sin theta -b cos the

1.	If a cos theta +b sin theta =c, then prove that a sin theta -b cos theta =+–√a^2+b^2–c^2
Answer» Given,\xa0acos{tex} \\theta{/tex}\xa0- b sin{tex} \\theta{/tex}\xa0= cSquaring on both sides(a cos{tex} \\theta{/tex}\xa0- b sin{tex} \\theta{/tex})2 = c2By Adding\xa0(a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2\xa0on both sides, we get\xa0(a cos{tex} \\theta{/tex}\xa0- b sin{tex} \\theta{/tex})2 + (a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2\xa0=c2 + (a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2(a2cos2{tex} \\theta{/tex}\xa0+ b2sin2{tex} \\theta{/tex}\xa0- 2ab sin{tex} \\theta{/tex}\xa0cos{tex} \\theta{/tex}) + (a2sin2{tex} \\theta{/tex}\xa0+ b2cos2{tex} \\theta{/tex}\xa0+ 2ab sin{tex} \\theta{/tex}\xa0cos{tex} \\theta{/tex})\xa0=c2 + (a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2a2(cos2{tex} \\theta{/tex}\xa0+ sin2{tex} \\theta{/tex}) + b2(sin2{tex} \\theta{/tex}\xa0+ cos2{tex} \\theta{/tex})=c2 + (a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2a2 + b2\xa0\xa0=\xa0c2 + (a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2{tex} \\Rightarrow{/tex}\xa0(a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex})2 = a2 + b2 - c2{tex} \\Rightarrow{/tex}\xa0a sin{tex} \\theta{/tex}\xa0+ b cos{tex} \\theta{/tex}\xa0=\xa0{tex} \\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}

If a cos theta +b sin theta =c, then prove that a sin theta -b cos theta =+–√a^2+b^2–c^2