Another trid identity problem

  • #1

Homework Statement


(1-cosx)/(1+cosx) = (cscx-cotx)^2


Homework Equations


cscx = 1/sinx, cotx = cosx/sinx, sin^2x + cos^2x = 1


The Attempt at a Solution



I have tried many different attempts, but I can't seem to make one side like the other. I took the (cscx-cotx)^2 and expanded it to csc^2x - 2cotxcscx+cot^2x = (1-2cosx+cos^2x)/sin^2x. Alternatively, (1-2cosx+cos^2x)/1-cos^2x

I've done every rearrangement I can think of, but I can't figure how to pull one side out of the other. Any help is much appreciated.
 

Answers and Replies

  • #2
378
2
Substitute cscx = 1/sinx, cotx = cosx/sinx in the right side and get only sin and cos equation on the right side.


I see you already done that:
(1-2cosx+cos^2x)/1-cos^2x = ((1-cosx)/(sinx))^2 is enough


Left side:
(1-cos.x)/(1+cosx)
.. looks like multiplying both num and den by 1-cosx would help
 
  • #3
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
Try multiplying top and bottom of the LHS by (1-cos x).
 

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