# Prove that: Tan 20+4sin20=sqrt3 This is till date the hardest question i have dealt with.

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You should use the fact that `tan alpha = sin alpha/cos alpha` , hence `tan 20 = sin 20/cos 20` such that:

`sin 20/cos 20 + 4sin 20 = sqrt3`

`sin 20 + 2*2 sin 20*cos 20 = sqrt3*cos 20`

`sin 20 + 2 sin 40 = sqrt3*cos 20`

`sin 20 + sin 40 + sin 40 = sqrt3*cos 20`

You need to convert the sum `sin 20 + sin 40 ` into a product using the formula `sin a + sin b = 2 sin ((a+b)/2)*cos((a - b)/2)`

`2 sin ((20+40)/2)*cos ((20 - 40)/2) + sin 40 = sqrt3*cos 20 `

`2sin 30*cos(-10) + sin 40 = sqrt3*cos 20`

You need to remember that `sin 30 = 1/2` and `cos(-a) = cos a` such that:

`2*(1/2) cos 10 + sin 40 = sqrt3*cos 20`

`cos 10 + sin 40 = sqrt3*cos 20`

You may write `cos 10 = sin(90 - 10) = sin 80,` hence, substituting `sin 80` for `cos 10` yields:

`sin 80 + sin 40 = sqrt3*cos 20`

You need to convert the sum `sin 80 + sin 40` into a product such that:

`2 sin((80+40)/2)cos((80-40)/2) = sqrt3*cos 20`

`2sin 60*cos(-20) = sqrt3*cos 20`

You need to remember that `sin 60 = sqrt3/2` and`cos (-20) = cos 20` such that:

`2*(sqrt3/2)*cos 20 = sqrt3*cos 20`

You need to reduce by 2 such that:

`sqrt3*cos 20 = sqrt3*cos 20`

**Notice that the last line proves the truthfulness of the given relation `tan 20 + 4sin 20 = sqrt3` .**

Thanks.