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tan2 0^0tan4 0^0tan6 0^0tan8 0^0...

`tan2 0^0tan4 0^0tan6 0^0tan8 0^0`

A

1

B

2

C

3

D

`(sqrt(3))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`tan20^(@)tan40^(@)tan60^(@)tan80^(@)`
`=(sin20^(@)sin40^(@)sin80^(@)tan60^(@))/(cos20^(@)cos40^(@)cos80^(@))`
Here `N^(r)=(sin20^(@)sin40^(@)sin80^(@))`
`=(sin20^(@))/(2)(2sin40^(@)sin80^(@))`
`=(sin20^(@))/(2)(cos40^(@)-cos120^(@))`
`=(1)/(2)sin20^(@)(1-2sin^(2)20^(@)+(1)/(2))`
`=(1)/(2)sin20^(@)((3)/(2)-2sin^(2)20^(@))`
`=(sin60^(@))/(4)=(sqrt(3))/(8)`
Now, we take `D^(r)=cos20^(@)cos40^(@)cos80^(@)`
`=(sin2^(3)20^(@))/(2^(3)sin20^(@))`
`=(sin160^(@))/(8sin20^(@))`
`=(sin20^(@))/(8sin20^(@))=(1)/(8)`
Hence, `tan20^(@)tan40^(@)tan80^(@)=(sqrt(3)//8)/(1//8)`
Therefore, `tan20^(@)tan40^(@)tan60^(@)tan80^(@)`
`=sqrt(3).sqrt(3)=3`
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