[tex]\bf \cfrac{1}{1+cos(x)}-\cfrac{1}{1-cos(x)}
\\\\\\
\textit{let us take the LCD of }[1+cos(x)][1-cos(x)]\qquad thus
\\\\\\
\cfrac{[1-cos(x)]-[1+cos(x)]}{[1+cos(x)][1-cos(x)]}\implies
\cfrac{1-cos(x)-1-cos(x)}{[1+cos(x)][1-cos(x)]}
\\\\\\\cfrac{-2cos(x)}{[1+cos(x)][1-cos(x)]}\\\\
-----------------------------\\\\
\textit{now recall your }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\\\
[/tex]
[tex]\bf -----------------------------\\\\
\cfrac{-2cos(x)}{[1+cos(x)][1-cos(x)]}\implies \cfrac{-2cos(x)}{1^2-cos^2(x)}\\\\
-----------------------------\\\\
\textit{now recall your pythagorean identities}\\\\
sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-----------------------------\\\\
\cfrac{-2cos(x)}{1^2-cos^2(x)}\implies \cfrac{-2cos(x)}{sin^2(x)}\implies
\cfrac{-2cos(x)}{sin(x)sin(x)}
\\\\\\
-2\cfrac{cos(x)}{sin(x)}\cdot \cfrac{1}{sin(x)}\implies -2cot(x)csc(x)[/tex]
