MULTIPLE-CHOICE QUEST What is the $$ \cos $$ A? ecant (sec), cosecant (csc) and cotangent (cot) example I Trigonometry I Khan Academy tan Academy $$ \operatorname{sic} A=\frac{9 p p}{n y p}=\frac{12}{13} $$ $$ \frac{5}{13} $$ $$ \frac{12}{13} $$ $$ \frac{13}{12} $$ $$ \frac{13}{5} $$

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To find $$ \cos A $$ given that $$ \sec A = \frac{12}{13} $$, we can use the relationship between secant and cosine. The secant function is the reciprocal of the cosine function:

$$
\sec A = \frac{1}{\cos A}
$$

From this, we can express $$ \cos A $$ as:

$$
\cos A = \frac{1}{\sec A}
$$

Now, substituting the given value of $$ \sec A $$:

$$
\cos A = \frac{1}{\frac{12}{13}} = \frac{13}{12}
$$

Thus, the value of $$ \cos A $$ is $$ \frac{13}{12} $$.

Now, let's check the options provided:

1. $$ \frac{5}{13} $$
2. $$ \frac{12}{13} $$
3. $$ \frac{13}{12} $$  (This is our answer)
4. $$ \frac{13}{5} $$

The correct answer is $$ \frac{13}{12} $$.
The value of $$ \cos A $$ is $$ \frac{13}{12} $$.

MULTIPLE-CHOICE QUEST What is the \( \cos \) A? ecant (sec), cosecant (csc) and cotangent (cot) example I Trigonometry I Khan Academy tan Academy \( \operatorname{sic} A=\frac{9 p p}{n y p}=\frac{12}{13} \) \( \frac{5}{13} \) \( \frac{12}{13} \) \( \frac{13}{12} \) \( \frac{13}{5} \)

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