# Trig Identities

• Jan 22nd 2008, 04:22 PM
Universe
Trig Identities
I am trying to solve the next trig identity however I can't figure it out after I convert the "1"s into cos^2 + sin^2.
$\displaystyle \frac {1+cos x} {sin x} + \frac {sin x} {1+cos x} = \frac {2} {sin x}$
I was also wondering if there's any way to convert the two into 2(cos^2 + sin^2)?
Thanks,
• Jan 22nd 2008, 05:04 PM
Soroban
Hello, Universe!

Quote:

$\displaystyle \frac {1+\cos x} {\sin x} + \frac {\sin x} {1+\cos x} \:= \:\frac {2}{\sin x}$
Just add the two fractions . . .

$\displaystyle \frac{1+\cos x}{\sin x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}} + \frac{\sin x}{1 + \cos x}\cdot{\color{blue}\frac{\sin x}{\sin x}} \;\;=\;\;\frac{(1+\cos x)^2}{\sin x(1+\cos x)} + \frac{\sin^2x}{\sin x(1+\cos x)}$

. . $\displaystyle = \;\;\frac{1 + 2\cos x + \overbrace{\cos^2\!x + \sin^2x}^{\text{This is 1}}}{\sin(1+\cos x)} \;\;=\;\;\frac{2 + 2\cos x}{\sin x(1 + \cos x)}$

. . $\displaystyle = \;\;\frac{2(1+\cos x)}{\sin x(1+\cos x)} \;\;=\;\;\frac{2}{\sin x}$