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)
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.