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Question

If 3sinθ+4cosθ=5, then the value of 4sinθ3cosθ is :

A
5
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B
2
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C
1
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D
0
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Solution

The correct option is D 0
Given,

3sin(θ)+4cos(θ)=5

3sin(θ)=54cos(θ)

(3sin(θ))2=(54cos(θ))2

9sin2(θ)(54cos(θ))2=0

(54cos(θ))2+(1cos2(θ))9=0

1625cos2(θ)+40cos(θ)=0

let, cos(θ)=u

1625u2+40u=0

25u2+40u16=0

u1,2=40±02(25)u=402(25)

u=45

cos(θ)=45

sin(θ)=35

Now,

4sin(θ)3cos(θ)

=4×353×45

=125125

=0


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