Find the equations of the tangent and normal to each

Find the equations of the tangent and normal to each curve at the given parameter value. 1° $x=\ln(\cos t)+1,\ y=\tan t+\cot t,\ t=\dfrac\pi4$. 2° $x=\sin t,\ y=a^{t},\ t=0$.

Step 1: Part 1°: point and derivatives

At $t=\dfrac\pi4$: $x=\ln\dfrac1{\sqrt2}+1 = 1-\dfrac{\ln2}{2}$, $y=1+1=2$. Point $P=\left(1-\dfrac{\ln2}2,\,2\right)$.

\[ \dot x = -\tan t, \qquad \dot y = \sec^{2}t-\csc^{2}t. \tag{1} \]

Step 2: Part 1°: slope at $t=\pi/4$

\[ \dot x\!\left(\tfrac\pi4\right) = -1, \qquad \dot y\!\left(\tfrac\pi4\right) = 2-2 = 0 \;\Longrightarrow\; k_{t}=\frac{0}{-1}=0. \tag{2} \]

The tangent is horizontal; the normal is vertical.

\[\text{Tangent: } y=2; \qquad \text{Normal: } x = 1-\frac{\ln2}{2}\]

Step 3: Part 2°: point and derivatives

At $t=0$: $x=0$, $y=a^{0}=1$. Point $(0,1)$.

\[ \dot x = \cos t, \qquad \dot y = a^{t}\ln a, \qquad \dot x(0)=1,\ \ \dot y(0)=\ln a. \tag{3} \]

Step 4: Part 2°: slope and lines

\[ k_{t} = \ln a, \qquad k_{n} = -\frac{1}{\ln a}\ \ (a\neq1). \tag{4} \]

\[\text{Tangent: } y = x\ln a+1; \qquad \text{Normal: } y = 1-\frac{x}{\ln a}\]

Step 5: Verification

\[1^\circ\text{: } \dot y\!\left(\tfrac\pi4\right)=\sec^2\tfrac\pi4-\csc^2\tfrac\pi4=2-2=0\ \checkmark\]

\[2^\circ\text{: tangent and normal at } (0,1) \text{ have slopes } \ln a \text{ and } -1/\ln a,\text{ product } -1\ \checkmark\]