The correct option is D 1→q, 2→r, 3→s, 4→p
cosθ−sinθ=15⋯(1)
Squaring equation (1), we get
1−sin2θ=125
⇒sin2θ=2425
2→r
Now,
(cosθ+sinθ)2=(cosθ−sinθ)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
1→q
Adding equation (1) and (2), we get
2cosθ=85⇒cosθ=45
4→p
Now,
cos2θ=2cos2θ−1⇒cos2θ=2×1625−1⇒cos2θ=725
3→s
Hence,
1→q, 2→r, 3→s, 4→p