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Q.

If θ1 and θ2 are two values lying in  [0,2π] for which tan⁡θ=λ, then tan⁡θ12tan⁡θ22 is equal to

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a

0

b

-1

c

2

d

1

answer is B.

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Detailed Solution

tan⁡θ=λ  we get  2tan⁡θ/21−tan2⁡θ/2=λ or  λtan2⁡θ2+2tan⁡θ2−λ=0⇒ tan⁡θ12tan⁡θ22=−1
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