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| Question 173390:  Please help me verify this equation:
 cscx -  cscx
 ------  ------   =2sec^2x
 1+cscx  1-cscx
 
 yes, the cscx's are over 1+cscx and 1-cscx. Those are fractions.
 So far i have gotten stuck on this porblem. i have expanded csc x into 1/sinx in both parts of the left side of the equation which gets me:
 1/sinx   -  1/sinx
 --------   --------- = 2sec^2x
 1+1/sinx   1-1/sinx
 
 After that i tried to achieve a common denominator on the left side which is where i get lost. I'm not sure what step to do after that or how to solve it.
 
 
 Answer by gonzo(654)
      (Show Source): 
You can put this solution on YOUR website! good one. maybe i got the answer.
 you want to prove this equation is true.
 here goes:
 -----
 let c = csc(x)
 let s = sec(x)
 -----
 your equation becomes:
 
  -----
 if you multiply both sides of the equation by (1-c)*(1+c), you get:
 
  -----
 simplifying, this becomes:
 
  which becomes:
 
  -----
 this is where substitutions come in.
 you know that:
 
  you also know that:
 sin(x) = 1/csc(x)
 and that:
 cos(x) = 1/sec(x)
 substituting for sin(x) and cos(x), the equation becomes:
 
  if you multiply both sides of this equation by
  , the equation becomes: 
  subtract
  from both sides of this equation to get: 
  simplify this on the right hand side to become:
 
  divide both sides of this equation by
  to get: 
  -----
 this is the same as:
 
  and, since we originally let c = csc(x) and we let s = sec(x), this equation now becomes:
 
  -----
 we can now substitute for c^2 in the original equation we derived above, which was:
 
  simplifying this, it becomes:
 
  -----
 substituting for
  , that equation becomes: 
  -----
 if we multiply both sides of this equation by
  , we get: 
  this becomes:
 
  which becomes:
 
  if we multiply both sides of this equation by (-1), it becomes:
 
  which proves that the original equation of:
 
  is true.
 -----
 since we originally substituted c for csc(x), and s for sec(x), the original equation becomes:
 
  and the proven identify becomes:
 
  -----
 
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