## Problem Statement

Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.

\[
\left\{
\begin{array}{l}
x=2 \ln (\operatorname{ctg} t) + \operatorname{ctg} t \\
y=\operatorname{tg} t + \operatorname{ctg} t
\end{array}
\right.
\]

$t_{0}=\frac{\pi}{4}$
## Solution

Since $t_{0}=\frac{\pi}{4}$, then

$x_{0}=2 \ln \left(\operatorname{ctg} \frac{\pi}{4}\right)+\operatorname{ctg} \frac{\pi}{4}=2 \ln 1+1=1$

$y_{0}=\operatorname{tg} \frac{\pi}{4}+\operatorname{ctg} \frac{\pi}{4}=1+1=2$

Let's find the derivatives:

$x_{t}^{\prime}=(2 \ln (\operatorname{ctg} t)+\operatorname{ctg} t)^{\prime}=2 \cdot \frac{1}{\operatorname{ctg} t} \cdot \frac{-1}{\sin ^{2} t}+\frac{-1}{\sin ^{2} t}=-\frac{1}{\sin ^{2} t} \cdot(2 \operatorname{tg} t+1)$

$y_{t}^{\prime}=(\operatorname{tg} t+\operatorname{ctg} t)^{\prime}=\frac{1}{\cos ^{2} t}-\frac{1}{\sin ^{2} t}$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1}{\cos ^{2} t}-\frac{1}{\sin ^{2} t}\right) /\left(-\frac{1}{\sin ^{2} t} \cdot(2 \operatorname{tg} t+1)\right)=\left(\frac{1}{\sin ^{2} t}-\frac{1}{\cos ^{2} t}\right) \cdot \frac{\sin ^{2} t}{2 \operatorname{tg} t+1}=$

$=\frac{1-\operatorname{tg}^{2} t}{2 \operatorname{tg} t+1}$

Then:
$y_{0}^{\prime}=\frac{1-\operatorname{tg}^{2} \frac{\pi}{4}}{2 \operatorname{tg} \frac{\pi}{4}+1}=\frac{1-1^{2}}{2 \cdot 1+1}=0$

Equation of the tangent line:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$

$y-2=0 \cdot(x-1)$

$y=2$

Since $y_{0}^{\prime}=0$, the equation of the normal line:

$x=x_{0}$

$x=1$

## Problem Kuznetsov Differentiation 17-9