Answer:
[tex]\huge{\text{\sf sec(x)+tan(x)}}[/tex]
Explanation:
[tex]\sf Given: \dfrac{cos(x)}{1-sin(x)}[/tex]
Rationalizing the denominator.
To rationalize an fraction such as a/(b+c) multiply both numerator and denominator with (b-c). Such as: [tex]\sf \frac{a}{(b+c)} \cdot \frac{b-c}{b-c}[/tex]
So if you apply the same here:
[tex]\rightarrow \sf \dfrac{cos(x)}{1-sin(x)} \cdot \dfrac{1+sin(x)}{1+sin(x)}[/tex]
[tex]\rightarrow \sf \dfrac{cos(x)(1+sin(x))}{1^2-sin^2(x)}[/tex]
[tex]\rightarrow \sf \dfrac{cos(x)+cos(x)sin(x)}{1-sin^2(x)}[/tex]
we know 1 - sin^2(x) = cos^2(x)
[tex]\rightarrow \sf \dfrac{cos(x)+cos(x)sin(x)}{cos^2(x)}[/tex]
breakdown
[tex]\rightarrow \sf \dfrac{cos(x)}{cos^2(x)}+\dfrac{cos(x)sin(x)}{cos^2(x)}[/tex]
simplify
[tex]\rightarrow \sf \dfrac{1}{cos(x)}+\dfrac{sin(x)}{cos(x)}[/tex]
1/cos(x) = sec(x), sin(x)/cos(x) = tan(x)
[tex]\rightarrow \sf sec(x)+tan(x)[/tex]
