1. ## trig

we have 2 secx= cosx/1-sinx+cosx/1+sinx
we must prove that it is equal
2secx=cosx(1-sinx)+cosx(1+sinx)/1-sinx+1+sinx

which give me
cosx-cosxsinx+cosx+cosxsinx/sinx^2
from this point i am lost have i make mistake ??

2. Originally Posted by gregorio
we have 2 secx= cosx/1-sinx+cosx/1+sinx
we must prove that it is equal
2secx=cosx(1-sinx)+cosx(1+sinx)/1-sinx+1+sinx

which give me
cosx-cosxsinx+cosx+cosxsinx/sinx^2
from this point i am lost have i make mistake ??

3. Originally Posted by gregorio
we have 2 secx= cosx/1-sinx+cosx/1+sinx
we must prove that it is equal
2secx=cosx(1-sinx)+cosx(1+sinx)/1-sinx+1+sinx

which give me
cosx-cosxsinx+cosx+cosxsinx/sinx^2
from this point i am lost have i make mistake ??
When you combined the two fractions into only one fraction, the denominator should have been (1-sinX)(1+sinX). Not your 1-sinX+1+sinX.
Like,
1/2 +1/3 = (1*3 +1*2)/(2*3) = 5/6
Not: = (1*3 +1*2)/(2+3) = 5/5 = 1. -------No.

So your second line should have been:
2secx = cosx(1-sinx) +cosx(1+sinx) / (1-sinx)(1+sinx)

Let me finish it.
2secX = cosX(1-sinX +1+sinX) / 1 -sin^2(X)
2secX = cosX(2) / cos^2(X)
2secX = 2 / cosX
2secX = 2(1/cosX)
2secX = 2secX