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sqrt(3)cossec2 0^0-sec2 0^0...

`sqrt(3)cossec2 0^0-sec2 0^0`

A

2

B

`sin20^(@)cosec40^(@)`

C

4

D

`4sin20^(@)cosec40^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
`sqrt(3)cosec20^(@)-sec20^(@)=(tan60^(@))/(sin20^(@))-(1)/(cos20^@)`
`=(sin60^(@)cos20^(@)-sin20^(@)cos60^(@))/(cos60^(@)sin20^(@)cos20^(@))`
`=(sin40^(@))/(cos60^(@)sin20^(@)cos20^(@))`
`=(2sin20^(@)cos20^(@))/((1)/(2)(sin20^(@)cos20^(@)))=4`
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