Complete the proof of the identity by choosing the rule that justifies each step.

[tex]\cos^2 x (1 + \tan^2 x) = 1[/tex]

\begin{tabular}{ll}
\textbf{Statement} & \textbf{Rule} \\
\hline
[tex]$\cos^2 x (1 + \tan^2 x)$[/tex] & Given \\
[tex]$=\cos^2 x (\sec^2 x)$[/tex] & [tex]\sec^2 x = 1 + \tan^2 x[/tex] \\
[tex]$=\cos^2 x \left(\frac{1}{\cos^2 x}\right)$[/tex] & [tex]\sec x = \frac{1}{\cos x}[/tex] \\
[tex]$=1$[/tex] & Simplification \\
\end{tabular}



Answer :

To prove the identity [tex]\(\cos^2 x\left(1 + \tan^2 x\right) = 1\)[/tex], we need to fill in the rules that justify each transformation step. Here's a detailed step-by-step solution:

1. Starting Expression:
[tex]\[ \cos^2 x(1 + \tan^2 x) \][/tex]

2. Step 1 (Using a Trigonometric Identity):
[tex]\[ \cos^2 x(1 + \tan^2 x) = \cos^2 x(\sec^2 x) \][/tex]
Rule: This step uses the trigonometric identity [tex]\(1 + \tan^2 x = \sec^2 x\)[/tex].

3. Step 2 (Definition of Secant):
[tex]\[ \cos^2 x(\sec^2 x) = \cos^2 x\left(\frac{1}{\cos^2 x}\right) \][/tex]
Rule: This step uses the definition of secant, [tex]\(\sec x = \frac{1}{\cos x}\)[/tex], thus [tex]\(\sec^2 x = \frac{1}{\cos^2 x}\)[/tex].

4. Step 3 (Simplification):
[tex]\[ \cos^2 x\left(\frac{1}{\cos^2 x}\right) = 1 \][/tex]
Rule: This step simplifies [tex]\(\cos^2 x \cdot \frac{1}{\cos^2 x} = 1\)[/tex] since the product of a term and its reciprocal is 1.

Combining these steps, we have:
[tex]\[ \cos^2 x(1 + \tan^2 x) = \cos^2 x(\sec^2 x) \quad \text{(Trigonometric Identity)} \][/tex]
[tex]\[ = \cos^2 x\left(\frac{1}{\cos^2 x}\right) \quad \text{(Definition of Secant)} \][/tex]
[tex]\[ = 1 \quad \text{(Simplification)} \][/tex]

So, the rules justifying each step are:
\begin{tabular}{ll}
[tex]$\cos ^2 x\left(1+\tan ^2 x\right)$[/tex] & Initial Expression \\
[tex]$=\cos ^2 x\left(\sec ^2 x\right)$[/tex] & Rule 1: Trigonometric Identity \\
[tex]$=\cos ^2 x\left(\frac{1}{\cos ^2 x}\right)$[/tex] & Rule 2: Definition of Secant \\
[tex]$=1$[/tex] & Rule 3: Simplification of Multiplication \\
\end{tabular}

Thus, the proof is complete.