Given, 3sinθ−5cosθ=3 Squaring on both side, we get (3sinθ−5cosθ)2=9 ⇒9sin2θ−30sinθcosθ+25cos2θ=9 ......(1) To find: 5sinθ+3cosθ Consider, 5sinθ+3cosθ=x Squaring on both side, we get (5sinθ+3cosθ)2=x2 ⇒25sin2θ+30sinθcosθ+9cos2θ=x2 .......(2) Adding (1) and(2), we get ⇒34sin2θ+34cos2θ=x2+9 ⇒34(sin2θ+cos2θ)=x2+9 ⇒34=x2+9 ⇒x2=25 ⇒x=±5 5sinθ+3cosθ=±5 Hence, If 3sinθ−5cosθ=3, then the value of 5sinθ+3cosθ is ±5