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Katelyn45 · Mar 29, 2011, 02:21 PM

ebaines · Mar 29, 2011, 02:40 PM

I'm guessing what you meant to write is this:

$<br /> \frac {1 + \tan x}{\sin x + \cos x} = \sec x<br />$

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

$<br /> \frac {1+\frac {\sin x} {\cos x}} {\sin x + \cos x}<br />$