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a
2(p2−q2)
b
2(p2+q2)
c
(p2+q2)
d
3(p2+q2)
answer is B.
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Detailed Solution
From given relation sinθp=cosθq=sin2θ+cos2θp2+q2=1p2+q2Now putting for p and q in the L.H.S. we have L.H.S.=(p2+q2)sinθsinϕ−cosθcosϕ=(p2+q2)sin(θ−ϕ)sinϕcosϕ=(p2+q2).2sin(3ϕ−ϕ)sin2ϕ=2p2−q2∵θ=3ϕ