Question

Which of the following would be an acceptable first step in simplifying the expression sinx/1-sinx

Answer 1

$\bf \textit{difference of squares} \\\\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b) \\\\\\ \textit{also recall that }sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -------------------------------$

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

Answer 2

Answer: D

Step-by-step explanation: