PLZ NEED HLP URGENT.

I kid, I kid. I had this question in my homework last week and tried probably 5 times and then wrote it off as a loss. Now it's come back around to abuse me on a take home quiz. So I've just now tried it another 7 or 8 times and about to pull my hair out. I'm supposed to pick either the left side or the right side, and using trig identities, work through the problem until the 2 sides are the same. I've tried both side and gotten nowhere.

 

Your first line is ok.
Don't expand the denominator.
Expand the nominator, as you have done on the second line.
Your result of 2 + 2sin is correct.
The identity should be obvious from there.

 

Note that this identity is not true under two conditions.

 

\( \frac{cos(x)}{1 + sin(x)} + \frac{1 + sin(x)}{cos{x}} = \)

\( \frac{cos^2(x) + [1 + sin(x)]^2}{[cos(x)][1 + sin(x)]} = \)

\( \frac{[cos^2(x) + sin^2(x)] + [1 + 2sin(x)]}{[cos(x)][1 + sin(x)]} = \)

\( \frac{2 + 2sin(x)}{[cos(x)][1 + sin(x)]} = \)

\( \frac{(2)(1 + sin(x))}{[cos(x)][1 + sin(x)]} = 2sec(x) \)

 

DOH! So simple. Thank you MrChips!

Edit: and amilton too!

 

\( {f:x \rightarrow 2sec(x) x\epsilon R\ x \neq (2n + 1) 90^{o} n \epsilon Z} \)