[tex]\bf \cfrac{1}{1-sin(x)}+\cfrac{1}{1+sin(x)}=\cfrac{2}{cos^2(x)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the LCD of [1-sin(x)][1+sin(x)]}}{\cfrac{[1+sin(x)]1~~+~~[1-sin(x)]1}{\underset{\textit{difference of squares}}{[1-sin(x)][1+sin(x)]}}} \\\\\\ \cfrac{1+sin(x)+1-sin(x)}{1^2-sin^2(x)}\implies \cfrac{1+sin(x)+1-sin(x)}{1-sin^2(x)}[/tex]
recall that 1 - sin²(θ) = cos²(θ).
