Question

How do I verify the identity: sec(t)+1/tan(t)=tan(t)/sec(t)-1

Answer 1

$\bf \textit{Pythagorean Identities} \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1 \\ \quad \\ 1+cot^2(\theta)=csc^2(\theta) \\ \quad \\ \boxed{1+tan^2(\theta)=sec^2(\theta)}\implies tan^2(\theta)=sec^2(\theta)-1\\\\ -----------------------------\\\\ \cfrac{sec(x)+1}{tan(x)}=\cfrac{tan(x)}{sec(x)-1}\\\\ -----------------------------$

$\bf \cfrac{sec(x)+1}{tan(x)}\cdot \cfrac{sec(x)-1}{sec(x)-1}\impliedby \textit{using the conjugate}\\\\ -----------------------------\\\\ recall\qquad \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ thus\qquad \cfrac{sec^2(x)-1^2}{tan(x)sec(x)-1}\implies \cfrac{sec^2(x)-1}{tan(x)sec(x)-1} \\\\\\ \cfrac{\underline{tan^2(x)}}{\underline{tan(x)} sec(x)-1}\implies \cfrac{tan(x)}{sec(x)-1}$