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Pre Calculus

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Prove that 2/square root 3cosx-sinx=sec(pi/6-x)

  • Pre Calculus -

    work on the right side

    RS = 1/cos(π/6 - x)
    = 1/)cosπ/6cosx + sinπ/6sinx)
    = 1/(√3/2cosx + 1/2sinx)
    = 2/(√3cosx + sinx)

    did you make a typing error?
    I tested your equation for some value of x, it did not work out, mine did.

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