Find the equation of the tangent line Find the equation of the line tangent to $f(x) = \tan x$ at $x = \frac{\pi}{4}$.
I'm trying to incorporate the slope point formula using $f'(x) = \sec^2 x$ but I'm nowhere near!
 A: At $x = \pi/4,\;$ $f(x) =y = \tan(\pi/4) = 1$. 
So $(\pi/4, 1)$  is the point of tangency of $f(x)$ and the desired line, which means it is certainly a point on the desired line.
Now, $f'(x) = \sec^2(x)$. At $x = \pi/4$, $$f'(x) = \sec^2(\pi/4) = \frac{1}{\cos^2(\pi/4)} = \left(\sqrt 2\right)^2 = 2$$ So the slope of the tangent line at the given point is $m = 2$.
Now use the point-slope form of an equation of a line, since you have needed point $(\pi/4, 1) = (a, b)$ and the slope $m = 2$:
$$(y - a) = m(x - b)$$
$$\begin{align} y - 1 = 2(x - \pi/4) & \iff y = 2x + (1 - \pi/2) \\ \\ &\iff y= 2x + \frac{2-\pi}{2}\end{align}$$
A: HINT:
We have $$f'\left(\frac\pi4\right)=\frac{y-f\left(\frac\pi4\right)}{x-\frac\pi4}$$
A: The tangent line is described by the equation: $$y=mx+b.$$
And we know that the slope of a tangent line to the graph of a function $f(x)$ at a point $(x_0, f(x_0))$ is the derivative of $f(x)$ evaluated at $x_0$, that is: $$m=f'(x_0).$$
So let's find it! $$m=\dfrac{\mathrm d}{\mathrm dx}\tan(x)\left|\right._{x={\pi}/{4}}=\sec^2\left(\tfrac{\pi}{4}\right)=2.$$
This tangent line passes through the point $\big(\pi/4, \tan(\pi/4)\big)=(\pi/4,1)$. By the slope-point formula: $$y-y_0=m(x-x_0)\Rightarrow y-1=2(x-\pi/4).$$
Rearranging we get: $$y=2x+\dfrac{2-\pi}2$$
which is our desired equation.
A: Equation of tangent at $x=x_0$ is given by $y-f(x_0)=f'(x_0)(x-x_0)$
Now put $x=\frac{\pi}{4}$ 
