SOLUTION: How does one prove that {{{(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cosA}}}?

Algebra ->  Trigonometry-basics -> SOLUTION: How does one prove that {{{(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cosA}}}?      Log On


   

Question 1114647: How does one prove that %28tanA%2BsecA-1%29%2F%28tanA-secA%2B1%29=%281%2BsinA%29%2FcosA?
Found 2 solutions by stanbon, josgarithmetic:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How does one prove that (tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cosA
-------
Cross-multiply to get::
cos(A)(tan(A)+sec(A)-1) = (1+sin(A))(tan(A)-sec(A)+1)
-------------
sin(A) +1 -cos(A) = tanA -secA +1 + sin^2A/cos(A) - tan(A) + sin(A)
---------
-cos(A) = -sec(A) + sin^2(A)/cos(A)
-----
-cos^2(A) = -1 + sin^2(A)
-----
sin^2(A)+cos^2(A) - 1 = 0
---
0 = 0
-----------
Cheers,
Stan H.
-----------

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the left side.

%28tanA%2B%28secA-1%29%29%2F%28tanA-%28secA-1%29%29

Try multiplying by %28tanA%2B%28secA-1%29%29%2F%28tanA%2B%28secA-1%29%29.

%28%28tanA%2B%28secA-1%29%29%5E2%29%2F%28%28tanA%29%5E2-%28secA-1%29%5E2%29

Just a start. I have not gone any further with this. Maybe you can and will finish.