Posted by **Nathan** on Thursday, April 22, 2010 at 1:27pm.

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

- Pre Calculus -
**Reiny**, Thursday, April 22, 2010 at 1:48pm
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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