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a
tanαtan2αtan3α=tan3α−tan2α−tanα
b
cosecα=cosec2α+cosec4α
c
cosα+cos2α+cos3α=1/2
d
8cosαcos2αcos4α=1
answer is A.
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Detailed Solution
1 tanαtan2αtan3α=tan3α−tan2α−tanα always holds good (∵tan2α=tan(3α−α))2 R.H.S. =sin4α+sin2αsin2αsin4α =2sin3αcosαsin2αsin4α=2sin4α⋅cosαsin2α⋅sin4α=1sinα=cosecα( using π/7=α) Hence, (2) is correct. Also cos2α=cos2π7=−cosπ−5π7=−cos5π7=−cos5α3 cosα+cos3α+cos5α=sin3αsinαcos3α=sin6α2sinα=sin6π72sinπ7=sinπ−π72sinπ7=124 8cosαcos2αcos4α=sin8αsinα=sin8π7sinπ7=−sinπ7sinπ7=−1