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tan 6^@ . Tan 42^@. Tan 66^@ . Tan 78^@...

`tan 6^@ . Tan 42^@. Tan 66^@ . Tan 78^@ ` is

A

1

B

`1//2`

C

`1//4`

D

`1//8`

Text Solution

Verified by Experts

The correct Answer is:
A
We have
`tan6^(@) tan 42^(@) tan66^(@)tan78^(@)`
`=tan 6^(@)tan(60^(@)-18^(@))tan(60^(@)+6^(@))tan(60^(@)+18^(@))`
`=([tan6^(@)tan(60^(@)+6^(@))tan18^(@)tan(60^(@)-18^(@))tan(60^(@)+18^(@))])/(tan18^(@))`
`=(tan6^(@)tan(60^(@)+6^(@))tan(3xx18^(@)))/(tan18^(@))`
`=(tan6^(@)tan(60^(@)-6^(@))tan(60^(@)+6^(@)))/(tan18^(@))`
`=(tan18^(@))/(tan18^(@))=1`
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