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Question

If θ1 and θ2 are two value lying on [0, 2π] for which tanθ = λ then tan θ12.tan θ22 is


A

0

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B

-1

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C

2

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D

1

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Solution

The correct option is B

-1


tanθ = λ2tanθ21tan2θ2 = λ

λ tan2 θ2 + 2tan θ2 - λ = 0

⇒ tan θ12 tan θ22 = -1


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