Question

Please answer question number 4

Answer 1

Answer:

acosθ+bsinθ=4 --- ( 1 )

⇒ asinθ−bcosθ=3 --- ( 2 )

→ now, squaring and adding ( 1 ) and ( 2 )

∴ (acosθ+bsinθ)2+(asinθ−bcosθ)2=42+32

⇒ a2cos2θ+2ab.sinθcosθ+b2sin2θ+a2sin2θ−2ab.sinθcosθ+b2cos2θ=16+9

⇒ a2(sin2θ+cos2θ)+b2(sin2θ+cos2θ)=25

∴ a2+b2=25 [∵sin2x+cos2x=1]