Question

find an equation of the tangent line to the curve at the given point.
y = \frac{1 + \sin(x)}{\cos(x)}, (\pi, -1)

Explanation:

Step1: Differentiate y using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 1+\sin(x)$ and $v=\cos(x)$. So, $u^\prime=\cos(x)$ and $v^\prime=-\sin(x)$. Then $y^\prime=\frac{\cos(x)\cdot\cos(x)-(1 + \sin(x))\cdot(-\sin(x))}{\cos^{2}(x)}=\frac{\cos^{2}(x)+\sin(x)+\sin^{2}(x)}{\cos^{2}(x)}$. Since $\sin^{2}(x)+\cos^{2}(x) = 1$, we have $y^\prime=\frac{1+\sin(x)}{\cos^{2}(x)}$.

Step2: Find the slope of the tangent line at $x = \pi$

Substitute $x=\pi$ into $y^\prime$. When $x = \pi$, $\sin(\pi)=0$ and $\cos(\pi)=-1$. Then $y^\prime|_{x = \pi}=\frac{1+0}{(-1)^{2}}=1$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(\pi,-1)$ and $m = 1$. So, $y+1=1\cdot(x - \pi)$.

Step4: Simplify the equation

$y=x-\pi - 1$.

Answer:

$y=x-\pi - 1$