Asked by Nathan

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

Answers

Answered by Reiny
work on the right side

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

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