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Question

If tan θ = pq then psinθqcosθpsinθ+qcosθ=

A
p2q2p2+q2
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B
q2p2q2+p2
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C
p2+q2p2q2
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D
1
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Solution

The correct option is A p2q2p2+q2
Divide both numerator and denominator by cosΘ.
psinθqcosθpsinθ+qcosθ = ptanθqptanθ+q
=
p(pq)qp(pq)+q = p2q2p2+q2

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