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(2cos 40^0-cos 20^0)/(sin 20^0)=...

`(2cos 40^0-cos 20^0)/(sin 20^0)=`

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(2cos40^(@)-cos20^(@))/(sin20^(@))

(2cos40^(@)-cos20^(@))/(sin20^(@))

Knowledge Check

  • (2cos40^(@)-cos20^(@))/(sin20^(@))=?

    A
    0
    B
    `sqrt(3)`
    C
    `-1`
    D
    2

  • What is the vlaue of (cos40^(@) - cos140^(@))//(sin80^(@)+sin20^(@))

    A
    `2sqrt3`
    B
    `2sqrt3`
    C
    `1sqrt3`
    D
    `sqrt3`

  • What is the value of ((cos40^(@)-cos140^(@)))/((sin80^(@)+sin20^(@))) ?

    A
    `2sqrt(3)`
    B
    `(2)/(sqrt(3))`
    C
    `(1)/(sqrt(3))`
    D
    `sqrt(3)`

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