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

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

A

4

B

2

C

1

D

0

Text Solution

Verified by Experts

The correct Answer is:
A
`sqrt(3)"cosec " 20^(@)-sec20^(@)`
`=(sqrt3)/(sin20^(@))-(1)/(cos20^(@))=(sqrt(3)cos20^(@)-sin20^(@))/(sin20^(@)cos20^(@))`
`=(4*(1)/(2)(sqrt(3)cos20^(@)-sin20^(@)))/(2sin20^(@)cos20^(@))`
`=((sqrt(3))/(2)cos20^(@)-(1)/(2)sin20^(@))((4)/(2sin20^(@)cos20^(@)))`
`=(sin60^(@)cos20^(@)-cos60^(@)sin20^(@))((4)/(sin40^(@)))`
`=sin40^(@)(4)/(sin40^(@))=4`
`(becausesin(A-B)=sinAcosB-cosAsinB)`
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