For `alpha=pi/7` which of the following hold (s) good?

Options

A. `tan alpha tan 2alpha tan 3alpha=tan 3alpha-tan 2alpha-tan alpha`
B. `co sec alpha=co sec 2alpha+ co sec 4alpha`.
C. `cos alpha-cos 2alpha+cos 3 alpha=1//2`
D. `8cos alpha cos 2 alpha cos 4alpha=1`

Answer

A, B, C

Explanation

(1) `tan alpha tan 2 alpha tan 3alpha=tan 3alpha-tan 2alpha-tan alpha`
always holds good, `(because tan 2alpha=tan(3alpha-alpha))`
(2) `RHS=(sin4alpha+sin2alpha)/(sin2alphasin4alpha)`
`=(2sin3alpha cos alpha)/(sin 2alpha sin 4alpha)=(2sin4alpha cos alpha)/(sin 2alpha.sin4alpha)`
`=(1)/(sin alpha)=co sec alpha` (using `pi//7=alpha`)
Hence, (2) is correct,
Also `cos 2alpha=cos((2pi)/(7))=-cos(pi-(5pi)/(7))`
`=-cos((5pi)/(7))=-cos 5alpha`.
(3) `cos alpha+cos 3alpha+cos5alpha=(sin3alpha)/(sinalpha)cos3alpha`
`=(sin6alpha)/(2sinalpha)=(sin((6pi)/(7)))/(2sin((pi)/(7)))`
`=(sin(pi-(pi)/(7)))/(2sin((pi)/(7)))=(1)/(2)`.
(4) `8cos alpha cos 2alpha cos 4alpha=(sin8alpha)/(sinalpha)=(sin((8pi)/(7)))/(sin((pi)/(7)))=(sin((pi)/(7)))/(sin((pi)/(7)))=-1`