`sin20^@(4+sec20^@)`
`=sin20^@(4+1/cos20^@)`
`=(4sin20^@cos20^@+sin20^@)/(cos20^@)`
`=(2sin40^@+sin20^@)/(cos20^@)`
`=(2sin(60-20)^@+sin20^@)/(cos20^@)`
`=(2sin60^@cos20^@-2cos60^@sin20^@+sin20^@)/(cos20^@)`
`=(2(sqrt3/2)cos20^@-2(1/2)sin20^@+sin20^@)/(cos20^@)`
`=(sqrt3cos20^@-sin20^@+sin20^@)/(cos20^@)`
`=sqrt3`
So, `sin20^@(4+sec20^@) = sqrt3`
