SOLUTION: Prove that : 1. (cscA-1)/(cscA+1) = (1-sinA)/(1+sinA) 2. (secA/cscA)+(sinA/cosA) = 2tanA 3. (1+sinA)/(1-sinA) = (cscA+1)/(cscA-1) 4. sinA/(sinA-cosA) = 1/(1-cotA) 5. (tanA+c

Algebra ->  Trigonometry-basics -> SOLUTION: Prove that : 1. (cscA-1)/(cscA+1) = (1-sinA)/(1+sinA) 2. (secA/cscA)+(sinA/cosA) = 2tanA 3. (1+sinA)/(1-sinA) = (cscA+1)/(cscA-1) 4. sinA/(sinA-cosA) = 1/(1-cotA) 5. (tanA+c      Log On


   

Question 549270: Prove that :
1. (cscA-1)/(cscA+1) = (1-sinA)/(1+sinA)
2. (secA/cscA)+(sinA/cosA) = 2tanA
3. (1+sinA)/(1-sinA) = (cscA+1)/(cscA-1)
4. sinA/(sinA-cosA) = 1/(1-cotA)
5. (tanA+cotA)/tanA = csc^2 A
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Prove that :
1. %28cscA-1%29%2F%28cscA%2B1%29 = %281-sinA%29%2F%281%2BsinA%29

  %28cscA-1%29÷%28cscA%2B1%29

%28+1%2FsinA-1+%29÷%28+1%2FsinA%2B1%29

%28+1%2FsinA-sinA%2FsinA+%29÷%28+1%2FsinA%2BsinA%2FsinA%29

+%281-sinA%29%2FsinA+÷+%281%2BsinA%29%2FsinA

+%281-sinA%29%2FsinA+×+sinA%2F%281%2BsinA%29

+%281-sinA%29%2Fcross%28sinA%29+×+cross%28sinA%29%2F%281%2BsinA%29

%281-sinA%29%2F%281%2BsinA%29


2. %28secA%2FcscA%29 + %28sinA%2FcosA%29 = 2tanA

   secA÷cscA + %28sinA%2FcosA%29

   1%2FcosA÷1%2FsinA%29 + %28sinA%2FcosA%29
    
   1%2FcosA×sinA%2F1%29 + %28sinA%2FcosA%29

   %28sinA%2FcosA%29 + %28sinA%2FcosA%29

   tanA + tanA

   2tanA


3. %281%2BsinA%29%2F%281-sinA%29 = %28cscA%2B1%29%2F%28cscA-1%29

You can do that one by substituting 1%2FsinA for cscA
It's similar to the first one.

4. sinA%2F%28sinA-cosA%29 = 1%2F%281-cotA%29
                    = 1÷1-cotA
                    = 1÷1-cosA%2FsinA 
                    = 1÷sinA%2FsinA-cosA%2FsinA
                    = 1÷%28sinA-cosA%29%2FsinA
                    = 1×sinA%2F%28sinA-cosA%29                 
                    = sinA%2F%28sinA-cosA%29

5. %28tanA%2BcotA%29%2FtanA = csc%5E2A
   tanA%2FtanA%2BcotA%2FtanA 
   1%2BcotA%2FtanA 
   1 + cotA%2FtanA
   1 + cotA÷tanA
   1 + cotA÷1%2FcotA
   1 + cotA×cotA%2F1
   1 + cotA×cotA  
   1 + cot%5E2A
   csc%5E2A

Edwin