1.

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`.


Discussion

No Comment Found

Related InterviewSolutions