one would note, that, you never really asked anything above
but.. in case you meant as in proof on the identities
well, doing the left-hand-side [tex]\bf \cfrac{1}{1+sin(x)}\cdot \cfrac{1-sin(x)}{1-sin(x)}\implies \cfrac{1-sin(x)}{[1+sin(x)][1-sin(x)]}\\\\
-----------------------------\\\\
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)\\\\
and\qquad sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{1-sin(x)}{1^2-sin^2(x)}\implies \cfrac{1-sin(x)}{1-sin^2(x)}\implies \cfrac{1-sin(x)}{cos^2(x)}[/tex]
