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Correct Answer – A
We have `tan( theta + (pi)/(4))= 3tan 3 theta `
or `(1+ tan theta)/(1-tan theta) = 3 xx (3tan theta – tan^(3) theta)/(1-3tan^(2) theta)`
`rArr (1+t)/(1-t) = 3 ((3t-t^(3))/(1-3t^(2)))” “` (putting `t = tan theta`)
`or 3t^(4) – 6t^(2) + 8t- 1 =0`
Hence,
`S_1 `= sum of roots `=t_1 + t_2 + t_3 + t_4 =0`
`S_2 ` = sum of product of roots taken two at a time = -2
`S_3` = sum of product of roots taken three at time `= -8//3`
`S_4` = product of all roots `=-1//3`
`(1)/(t_1) + (1)/(t_2) + (1)/(t_3) + (1)/(t_4) = (sum t_1 t_2 t_3)/(t_2t_2t_3t_4) = (-8//3)/(-1//3) =8`
Correct Answer – D
We have `tan( theta + (pi)/(4))= 3tan 3 theta `
or `(1+ tan theta)/(1-tan theta) = 3 xx (3tan theta – tan^(3) theta)/(1-3tan^(2) theta)`
`rArr (1+t)/(1-t) = 3 ((3t-t^(3))/(1-3t^(2)))” “` (putting `t = tan theta`)
`or 3t^(4) – 6t^(2) + 8t- 1 =0`
Hence,
`S_1 `= sum of roots `=t_1 + t_2 + t_3 + t_4 =0`
`S_2 ` = sum of product of roots taken two at a time = -2
`S_3` = sum of product of roots taken three at time `= -8//3`
`S_4` = product of all roots `=-1//3`
`(1)/(t_1) + (1)/(t_2) + (1)/(t_3) + (1)/(t_4) = (sum t_1 t_2 t_3)/(t_2t_2t_3t_4) = (-8//3)/(-1//3) =8`