Answer
$y=12x+2-3\pi.$
Work Step by Step
$u=\tan{x}$; $\dfrac{du}{dx}=\sec^2{x}$
$\dfrac{dy}{du}=6u^2$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=6\sec^2{x}\tan^2{x}$
$y'$ evaluated at $\dfrac{\pi}{4}\rightarrow y'=6\sec^2{\dfrac{\pi}{4}}\tan^2{\dfrac{\pi}{4}}=12$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-2)=12(x-\dfrac{\pi}{4})\rightarrow y=12x+2-3\pi.$
A graphing calculator and a computer algebra system have been used to confirm these results.
