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tan20+4sin20 |
Answer» `sin20/cos20+4sin20` `(sin20+4sin20cos20)/cos20` `2sinAcosA=sin2A` `sin(C-D)=sinCcosD-cosCsinD` `sin(C+D)=sinCcosD+cosCsinD` `(sinn20+2sin40)/cos20` `(sin20+2sin(60-20))/cos20` `(sin20+2{sin60cos20-cos60sin20})/cos20` `(sin20+2sin60cos20-sin20)/cos20` `(2sin60cos20)/cos20` `2sin60` `2*sqrt3/2` `sqrt3`. |
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