[tex]\bf \textit{Pythagorean Identities}
\\ \quad \\
sin^2(\theta)+cos^2(\theta)=1
\\ \quad \\
1+cot^2(\theta)=csc^2(\theta)
\\ \quad \\
\boxed{1+tan^2(\theta)=sec^2(\theta)}\implies tan^2(\theta)=sec^2(\theta)-1\\\\
-----------------------------\\\\
\cfrac{sec(x)+1}{tan(x)}=\cfrac{tan(x)}{sec(x)-1}\\\\
-----------------------------\\\\[/tex]
[tex]\bf \cfrac{sec(x)+1}{tan(x)}\cdot \cfrac{sec(x)-1}{sec(x)-1}\impliedby \textit{using the conjugate}\\\\
-----------------------------\\\\
recall\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
thus\qquad \cfrac{sec^2(x)-1^2}{tan(x)sec(x)-1}\implies \cfrac{sec^2(x)-1}{tan(x)sec(x)-1}
\\\\\\
\cfrac{\underline{tan^2(x)}}{\underline{tan(x)} sec(x)-1}\implies \cfrac{tan(x)}{sec(x)-1}[/tex]
