if acosx-bsinx=c Prove that asinx+bcosx= sq rt a^2+b^2-c^2 - Maths - Some Applications of Trigonometry

Question

if acosx-bsinx=c Prove that asinx+bcosx= sq rt a^2+b^2-c^2

Gursheen kaur. · Accepted Answer

Given, acosx-bsinx=c (i) ${(a \cos x - b \sin x)}^{2} + {(a \sin x + b \cos x)}^{2} = a^{2} \cos^{2} x + b^{2} \sin^{2} x - 2 a b \cos x \sin x + a^{2} \sin^{2} x + b^{2} \cos^{2} x + 2 a b \cos x \sin x = a^{2} \cos^{2} x + a^{2} \sin^{2} x + b^{2} \sin^{2} x + b^{2} \cos^{2} x - 2 a b \cos x \sin x + 2 a b \cos x \sin x = a^{2} (\cos^{2} x + \sin^{2} x) + b^{2} (\sin^{2} x + \cos^{2} x) = a^{2} + b^{2} [∵ \sin^{2} θ + \cos^{2} θ = 1]$ ${(a \cos x - b \sin x)}^{2} + {(a \sin x + b \cos x)}^{2} = a^{2} + b^{2} \Rightarrow c^{2} + {(a \sin x + b \cos x)}^{2} = a^{2} + b^{2} [∵ o f (i)] \Rightarrow {(a \sin x + b \cos x)}^{2} = a^{2} + b^{2} - c^{2} \Rightarrow a \sin x + b \cos x = \sqrt{a^{2} + b^{2} - c^{2}}$

if acosx-bsinx=c. Prove that asinx+bcosx= sq.rt a^2+b^2-c^2.