To provide a step-by-step explanation and review of the proof, we need to focus on how to transition from step 1 to step 2:
[tex]\[
\begin{array}{|l|l|}
\hline
\text{Step} & \text{Statement} \\
\hline
1 & \tan \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos (x)}{1+\cos (x)}} \\
\hline
2 & \tan \left(\frac{x}{2}\right)=\left(\frac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)}}\right)\left(\frac{7}{\sqrt{1+\cos (x)}}\right) \\
\hline
\end{array}
\][/tex]
In step 1, we have:
[tex]\[
\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}
\][/tex]
Step 2 shows:
[tex]\[
\tan \left(\frac{x}{2}\right)=\left(\frac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)}}\right)\left(\frac{7}{\sqrt{1+\cos (x)}}\right)
\][/tex]
To complete the expression in step 2, we notice that the fraction inside the square root in step 1:
[tex]\[
\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}
\][/tex]
is split into two parts:
[tex]\[
\left(\frac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)}}\right)
\][/tex]
and another term:
[tex]\[
\left(\frac{7}{\sqrt{1+\cos (x)}}\right)
\][/tex]
To transition from step 1 to step 2, we multiply and divide by [tex]\(\sqrt{1+\cos(x)}\)[/tex]. Thus, we are effectively introducing the denominator [tex]\(\sqrt{1+\cos(x)}\)[/tex] again in both the numerator and denominator of the fraction.
The expression that completes the step is:
[tex]\[
1 + \cos(x)
\][/tex]
When we multiply the numerator and denominator in the fraction [tex]\(\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\)[/tex] by [tex]\(\sqrt{1+\cos(x)}\)[/tex], we obtain (after splitting appropriately):
[tex]\[
\tan \left(\frac{x}{2}\right) = \left( \frac{\sqrt{1-\cos(x)}}{\sqrt{1+\cos(x)}} \right) \left( \frac{\sqrt{1+\cos(x)}}{\sqrt{1+\cos(x)}} \right)
\][/tex]
Thus, incorporating the given expression correctly aligns the proof.
Therefore, the expression that will complete step 2 in the proof is:
[tex]\[
1 + \cos (x)
\][/tex]
