If a cos θ + b sin θ and a sin θ b cos θ = 3, then a2 + b2 =

Given: a cos θ + b sin θ = 4

Squaring both sides, we get


(a cos θ + b sin θ)2 = 42


a2 cos2 θ + b2 sin2 θ + 2ab sin θ cos θ = 16 ………(i)


and a sin θ – b cos θ = 3


Squaring both sides, we get


(a sin θ – b cos θ)2 = 32


a2 sin2 θ + b2 cos2 θ – 2ab sin θ cos θ = 9 ………(ii)


To find: a2 + b2


Adding (i) and (ii), we get


a2 cos2 θ + b2 sin2 θ + 2ab sin θ cos θ


+ a2 sin2 θ + b2 cos2 θ – 2ab sin θ cos θ = 16 + 9


a2 (sin2 θ + cos2 θ) + b2 (sin2 θ + cos2 θ) = 25


a2 + b2 = 25 [ sin2 θ + cos2 θ = 1]

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