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Q.

If α=π14 then the value of (tanαtan2α+tan2αtan4α+tan4αtanα) is

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a

2

b

1

c

12

d

13

answer is A.

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Detailed Solution

α+2α+4α=7α=π2

 tanαtan2α+tan2αtan4α+tanαtan4α=1

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