Question

3sin A + 5cos a = 5 then prove 3cosa-5sina=±3

Answer 1

Hey,

3sinA+5cosA=53sin⁡A+5cos⁡A=5 -(1)

5sinA−3cosA=m,5sin⁡A−3cos⁡A=m, (suppose) -(2)

Squaring and adding equations (1) and (2),

(3sinA+5cosA)2+(5sinA−3cosA)2=52+m2(3sin⁡A+5cos⁡A)2+(5sin⁡A−3cos⁡A)2=52+m2

∴9sin2A+2(3)(5)sinAcosA+25sin2A+25sin2A−2(3)(5)sinAcosA+9cos2A=25+m2∴9sin2⁡A+2(3)(5)sin⁡Acos⁡A+25sin2⁡A+25sin2⁡A−2(3)(5)sin⁡Acos⁡A+9cos2⁡A=25+m2

∴9+25=25+m2∴9+25=25+m2

(∵sin2x+cos2x=1∵sin2⁡x+cos2⁡x=1)

∴m=3∴m=3
Hope it helps u