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Review the proof.

\begin{tabular}{|l|l|}
\hline
Step & Statement \\
\hline
1 & [tex]$\tan \left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos(x)}{1 + \cos(x)}}$[/tex] \\
\hline
2 & [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] \\
\hline
3 & [tex]$\tan \left(\frac{x}{2}\right) = \frac{\sqrt{1 - \cos^2(x)}}{1 + \cos(x)}$[/tex] \\
\hline
4 & [tex]$\tan \left(\frac{x}{2}\right) = \frac{\sqrt{\sin^2(x)}}{1 + \cos(x)}$[/tex] \\
\hline
5 & [tex]$\tan \left(\frac{x}{2}\right) = \frac{\sin(x)}{1 + \cos(x)}$[/tex] \\
\hline
\end{tabular}

Which expression will complete step 2 in the proof?

A. [tex]$1 - \cos(x)$[/tex]
B. [tex]$1 + \cos(x)$[/tex]
C. [tex]$\sqrt{1 - \cos(x)}$[/tex]
D. [tex]$\sqrt{1 + \cos(x)}$[/tex]

Sagot :

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]