now, recall that the conjugate of a + b, is just a - b, therefore, the conjugate of 1 - sin(x) is just 1 + sin(x).
so hmmm let's multiply the left-side by the conjugate of the numerator then.
[tex]\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\[/tex]
[tex]\bf \cfrac{1-sin(x)}{cos(x)}=\cfrac{cos(x)}{1+sin(x)}
\\\\\\
\cfrac{1-sin(x)}{cos(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \cfrac{[1-sin(x)][1+sin(x)]}{cos(x)[1+sin(x)]}
\\\\\\
\cfrac{1^2-sin^2(x)}{cos(x)[1+sin(x)]}\implies \cfrac{\underline{cos^2(x)}}{\underline{cos(x)}[1+sin(x)]}\implies \cfrac{cos(x)}{1+sin(x)}[/tex]
