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Question

If sinθ+cosθ=p,sin3θ+cos3θ=q, then p(p23)=

A
q
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B
2q
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C
q
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D
2q
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Solution

The correct option is D 2q
Given, sinθ+cosθ=p, sin3θ+cos3θ=q

We know a3+b3=(a+b)(a2+b2ab)

Therefore, sin3θ+cos3θ=(sinθ+cosθ)(sin2θ+cos2θsinθcosθ)

q=p(1sinθcosθ)
qp=1sinθcosθ

sinθ+cosθ=p

1+2sinθcosθ=p2

2sinθcosθ=p21

sinθcosθ=p212

qp=[1(p21)2]

2q=p(2p2+1)

2q=p(3p2)

p(p23)=2q

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