a cos A -b sin A =c ,then prove a sin A+b cos A =√a^2+b^2-c^2

1.	a cos A -b sin A =c ,then prove a sin A+b cos A =√a^2+b^2-c^2
Answer» We have, {tex}asin\\theta+bcos\\theta=c{/tex}On squaring both sides, we get{tex}(asin\\theta+bcos\\theta)^2=c^2{/tex}(a sin θ)2\xa0+ (b cos θ)2\xa0+ 2(a sin θ) (b cos θ) = c2⇒ a2\xa0sin2\xa0θ + b2\xa0cos2\xa0θ + 2ab sin θ cos θ = c2⇒ a2(1 – cos2\xa0θ) + b2\xa0(1 – sin2\xa0θ) + 2 ab sin θ cos θ = c2 {tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ a2\xa0– a2\xa0cos2\xa0θ + b2\xa0– b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2⇒ –a2\xa0cos2\xa0θ – b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2\xa0– a2\xa0– b2 Taking Negative common,⇒ a2\xa0cos2\xa0θ + b2\xa0sin2\xa0θ – 2ab sin θ cos θ = a2\xa0+ b2\xa0– c2⇒ (a cos θ)2\xa0+ (b sin θ)2\xa0– 2(a cos θ) (b sin θ) = a2\xa0+ b2\xa0– c2⇒ {tex}(acos\\theta-bsin\\theta)^2=a^2+b^2-c^2{/tex}⇒{tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}\xa0Hence proved, {tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\sqrt{a^2+b^2-c^2}{/tex}

Discussion