Answer
$y=x-\pi-1$
Work Step by Step
Given:
$y=\frac{1+\sin x}{\cos x}$
$y=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$
$y=\sec x+\tan x$
Find the slope of the tangent line to the curve at $x=\pi$:
$\frac{dy}{dx}=\frac{d}{dx}(\sec x+\tan x)$
$\frac{dy}{dx}=\frac{d}{dx}(\sec x)+\frac{d}{dx}(\tan x)$
$\frac{dy}{dx}=\sec x\tan x+\sec^2 x$
$m=\frac{dy}{dx}|_{x=\pi}$
$m=\sec \pi\tan\pi+\sec^2\pi$
$m=-1\cdot 0+(-1)^2$
$m=0+1$
$m=1$
Find the equation of the tangent line:
$y-(-1)=1(x-\pi)$
$y+1=x-\pi$
$y=x-\pi-1$
Thus, the tangent line has the equation $y=x-\pi-1$.
