Question
Prove that 2/square root 3cosx-sinx=sec(pi/6-x)
Answers
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)
did you make a typing error?
I tested your equation for some value of x, it did not work out, mine did.
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.