The problem asks to find the equation of the tangent line $g$ to the graph of the function $y = \tan x$ at the point $(\frac{\pi}{4}, 1)$. There seems to be a typo in the coordinates of the point, and based on the equation $y = \tan x$, the point should be $(\frac{\pi}{4}, 1)$ since $\tan(\frac{\pi}{4}) = 1$. We will assume the point of tangency is $(\frac{\pi}{4}, 1)$.

1. Problem Description

The problem asks to find the equation of the tangent line $g$ to the graph of the function $y = \tan x$ at the point $(\frac{\pi}{4}, 1)$. There seems to be a typo in the coordinates of the point, and based on the equation $y = \tan x$, the point should be $(\frac{\pi}{4}, 1)$ since $\tan(\frac{\pi}{4}) = 1$. We will assume the point of tangency is $(\frac{\pi}{4}, 1)$.

2. Solution Steps

To find the equation of the tangent line, we need to find the slope of the tangent line and a point on the line. We are given the point $(\frac{\pi}{4}, 1)$.

The slope of the tangent line is the derivative of the function evaluated at $x = \frac{\pi}{4}$.

The derivative of $y = \tan x$ is $y' = \sec^2 x$.

We evaluate $y'$ at $x = \frac{\pi}{4}$:

$y'(\frac{\pi}{4}) = \sec^2(\frac{\pi}{4})$

Recall that $\sec x = \frac{1}{\cos x}$. Also, $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, so $\sec(\frac{\pi}{4}) = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Therefore, $\sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2$.

The slope of the tangent line at $x = \frac{\pi}{4}$ is $m = 2$.

Now we can use the point-slope form of a line to find the equation of the tangent line:

$y - y_1 = m(x - x_1)$

Here, $(x_1, y_1) = (\frac{\pi}{4}, 1)$ and $m = 2$.

$y - 1 = 2(x - \frac{\pi}{4})$

$y - 1 = 2x - \frac{\pi}{2}$

$y = 2x - \frac{\pi}{2} + 1$

3. Final Answer

The equation of the tangent line is $y = 2x - \frac{\pi}{2} + 1$.