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Question

If tanθ=pq and θ=3 ϕ(0<θ<π2), prove
that psinϕqcosϕ=2(p2+q2)

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Solution

From given relation
sinθp=cosθq=sin2θ+cos2θp2+q2=1p2+q2
Now putting for p in the q in the L.H.S., we have
L.H.S=(p2+q2)[sinθsinϕcosθcosϕ]=(p2+q2)sin(θϕ)sinϕcosϕ
=(p2+q2).2sin(3ϕϕ)sin2ϕ=p2+q2θ=3ϕ

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