Question

I need help solving this trig problem w/ identites.

Answer 1

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad sec(\theta)=\cfrac{1}{cos(\theta)} \\\\ -------------------------------$

$\bf \cfrac{1}{sin(x)-1}+\cfrac{1}{sin(x)+1}\implies \cfrac{sin(x)+1~~+~~sin(x)-1}{[sin(x)-1][sin(x)+1]} \\\\\\ \cfrac{2sin(x)}{\stackrel{\textit{difference of squares}}{sin^2(x)-1^2}}\implies \cfrac{2sin(x)}{sin^2(x)-1}\implies \cfrac{2sin(x)}{-[~1-sin^2(x)~]} \\\\\\ \cfrac{2sin(x)}{-[cos^2(x)]}\implies -2\cdot \cfrac{sin(x)}{cos(x)}\cdot \cfrac{1}{cos(x)}\implies -2tan(x)sec(x)$