Find the exact value of the trigonometric function of $$ \theta $$ from the given information. $$ \cos \theta=\frac{9}{41}, \theta \text { in quadrant I, find } \tan \theta $$

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To find $$\tan \theta$$ given that $$\cos \theta = \frac{9}{41}$$ and $$\theta$$ is in the first quadrant, we can use the Pythagorean identity:

$$
\sin^2 \theta + \cos^2 \theta = 1
$$

First, we can find $$\sin \theta$$ using the given value of $$\cos \theta$$:

$$
\sin^2 \theta + \left(\frac{9}{41}\right)^2 = 1
$$

Calculating $$\left(\frac{9}{41}\right)^2$$:

$$
\left(\frac{9}{41}\right)^2 = \frac{81}{1681}
$$

Now substituting this into the Pythagorean identity:

$$
\sin^2 \theta + \frac{81}{1681} = 1
$$

To isolate $$\sin^2 \theta$$, we subtract $$\frac{81}{1681}$$ from both sides:

$$
\sin^2 \theta = 1 - \frac{81}{1681}
$$

To perform the subtraction, we convert $$1$$ into a fraction with a denominator of $$1681$$:

$$
1 = \frac{1681}{1681}
$$

Now we can subtract:

$$
\sin^2 \theta = \frac{1681}{1681} - \frac{81}{1681} = \frac{1681 - 81}{1681} = \frac{1600}{1681}
$$

Next, we take the square root to find $$\sin \theta$$:

$$
\sin \theta = \sqrt{\frac{1600}{1681}} = \frac{\sqrt{1600}}{\sqrt{1681}} = \frac{40}{41}
$$

Since $$\theta$$ is in the first quadrant, $$\sin \theta$$ is positive.

Now we can find $$\tan \theta$$:

$$
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9}
$$

Thus, the exact value of $$\tan \theta$$ is:

$$
\boxed{\frac{40}{9}}
$$
The exact value of $$\tan \theta$$ is $$\frac{40}{9}$$.

Find the exact value of the trigonometric function of from the given information.

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