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Question

Question 11
If a sin θ+b cos θ=c, then prove that a cos θb sin θ=a2+b2c2

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Solution

Given that, a sin θ+b cos θ=c

On squaring both sides,

(a sin θ+b cos θ)2=c2

a2 sin2 θ+b2 cos2 θ+2ab sin θ.cos θ=c2

[(x+y)2=x2+2xy+y2]

a2(1cos2 θ)+b2(1sin2 θ)+2ab sin θ.cos θ=c2

[sin2 θ+cos2 θ=1]

a2a2 cos2 θ+b2b2 sin2 θ+2ab sin θ.cos θ=c2

a2+b2c2=a2 cos2 θ+b2 sin2 θ2ab sin θ.cos θ

(a2+b2c2)=(a cos θb sin θ)2[a2+b22ab=(ab)2]

(a cos θb sin θ)2=a2+b2c2

a cos θb sin θ=a2+b2c2

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