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Question

If cosθsinθ=15, where 0<θ<π2

List IList II(1)(cosθ+sinθ)2(p)45(2)sin2θ(q)710(3)cos2θ(r)2425(4)cosθ(s)725

Which of the following is the correct combination?

A
1s, 2p, 3r, 4q
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B
1p, 2s, 3r, 4p
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C
1q, 2s, 3r, 4p
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D
1q, 2r, 3s, 4p
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Solution

The correct option is D 1q, 2r, 3s, 4p
cosθsinθ=15(1)
Squaring equation (1), we get
1sin2θ=125
sin2θ=2425
2r

Now,
(cosθ+sinθ)2=(cosθsinθ)2+4sinθcosθ
=125+2sin2θ
=125+4825
=4925
(cosθ+sinθ)2=4925
(cosθ+sinθ)=±75
Since θ(0,π2)
cosθ+sinθ=75(2)
(cosθ+sinθ)2=710
1q

Adding equation (1) and (2), we get
2cosθ=85cosθ=45
4p

Now,
cos2θ=2cos2θ1cos2θ=2×16251cos2θ=725
3s

Hence,
1q, 2r, 3s, 4p

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