For α=π7 which of the following hold(s) good?

1. tan⁡αtan⁡2αtan⁡3α=tan⁡3α−tan⁡2α−tan⁡α   always holds good2.     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, (b) is correct.     Also,    3 .  cos⁡2α=cos⁡2π7=−cos⁡π−5π7=−cos⁡5π7=−cos⁡5α.     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