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Question

If 3sinθ+4cosθ=5, then find the value of 4sinθ3cosθ.

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Solution

We have,

3sinθ+4cosθ=5 ………. (1)

On squaring both sides, we get

(3sinθ+4cosθ)2=52

9sin2θ+16cos2θ+24sinθcosθ=25

9(1cos2θ)+16(1sin2θ)+12×2sinθcosθ=25

99cos2θ+1616sin2θ+12×2sinθcosθ=25

259cos2θ16sin2θ+12×2sinθcosθ=25

9cos2θ+16sin2θ12×2sinθcosθ=0

(3cosθ4sinθ)2=0

3cosθ4sinθ=0

Hence, the value is 0.


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