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Q.

For α=π/7 which of the following hold(s) good?

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a

tan⁡αtan⁡2αtan⁡3α=tan⁡3α−tan⁡2α−tan⁡α

b

cosec⁡α=cosec⁡2α+cosec⁡4α

c

cos⁡α+cos⁡2α+cos⁡3α=1/2

d

8cos⁡αcos⁡2αcos⁡4α=1

answer is A.

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Detailed Solution

1  tan⁡αtan⁡2αtan⁡3α=tan⁡3α−tan⁡2α−tan⁡α always holds good (∵tan⁡2α=tan⁡(3α−α))2   R.H.S. =sin⁡4α+sin⁡2αsin⁡2αsin⁡4α                    =2sin⁡3αcos⁡αsin⁡2αsin⁡4α=2sin⁡4α⋅cos⁡αsin⁡2α⋅sin⁡4α=1sin⁡α=cosec⁡α( using π/7=α)                   Hence, (2) is correct.                    Also cos⁡2α=cos⁡2π7=−cos⁡π−5π7=−cos⁡5π7=−cos⁡5α3   cos⁡α+cos⁡3α+cos⁡5α=sin⁡3αsin⁡αcos⁡3α=sin⁡6α2sin⁡α=sin⁡6π72sin⁡π7=sin⁡π−π72sin⁡π7=124  8cos⁡αcos⁡2αcos⁡4α=sin⁡8αsin⁡α=sin⁡8π7sin⁡π7=−sin⁡π7sin⁡π7=−1
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