If `a_(1),a_(2),a_(3),……a_(87),a_(88),a_(89)` are the arithmetic means between `1` and `89`, then `sum_(r=1)^(89)log(tan(a_(r ))^(@))` is equal to
A
`0`
B
`1`
C
`log_(2)3`
D
`log5`
Text Solution
Verified by Experts
The correct Answer is:
A
`(a)` As `1, a_(1),a_(2),a_(3),……….,a_(87),a_(88),a_(89),89` are in `A.P`. So, `a_(1)+a_(89)=a_(2)+a_(88)=……..=1+89=90` `:.sum_(r=1)^(89)log(tana_(r )^(@))` `=log(tana_(1)^(@)*tana_(2)^(@)……..tana_(88)^(@)*tana_(89)^(@))` `=log1=0`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )
