ebaines
Mar 29, 2011, 02:40 PM
I'm guessing what you meant to write is this:
\frac {1 + \tan x}{\sin x + \cos x} = \sec x
Rewrite the \tan x and \sec x functions in terms of their primitive sin x and \cos x equivalents, and it pops right out.
Katelyn45
Mar 29, 2011, 03:02 PM
Then I have (1+(cosx/sinx))/Sinx+cosx
ebaines
Mar 30, 2011, 05:47 AM
Remember that tan(x) = sin(x)/cos(x). So you have
\frac {1+\frac {\sin x} {\cos x}} {\sin x + \cos x}