Answer
Tangent line: $\quad y=-\displaystyle \frac{1}{2}x-\frac{1}{2}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-\frac{\pi}{4}}=\frac{1}{4}$
Work Step by Step
The point $P$ on the curve corresponding to $t=-\displaystyle \frac{\pi}{4}$ is
$\left\{\begin{array}{l}
x=\sec^{2}(-\frac{\pi}{4})-1=1\\
\\
y=\tan(-\frac{\pi}{4})=-1
\end{array}\right.\quad, P(1,-1)$
The slope of the tangent line at $t=-\displaystyle \frac{\pi}{4}$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=-\frac{\pi}{4}}$
$\displaystyle \left\{\begin{array}{l}
\frac{dx}{dt}=2\sec^{2}t\tan t,\\
\\
\frac{dy}{dt}=\sec^{2}t
\end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\sec^{2}t}{2\sec^{2}t\tan t}=\frac{\mathrm{l}}{2\tan t}=\frac{1}{2}\cot t$
$\left. \displaystyle \frac{dy}{dx} \right|_{t=-\frac{\pi}{4}}=\frac{1}{2}\cot(-\frac{\pi}{4})=-\frac{1}{2}$
Use the point-slope equation for the tangent line.
$y+1=-\displaystyle \frac{1}{2}(x-1)$
$y=-\displaystyle \frac{1}{2}x+\frac{1}{2}-1$
$y=-\displaystyle \frac{1}{2}x-\frac{1}{2}$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-\frac{\pi}{4}}$,
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\frac{d}{dt}[\frac{1}{2}\cot t]}{2\sec^{2}t\tan t}=\frac{\frac{1}{2}(-\csc^{2}t)}{2\sec^{2}t\tan t}=-\frac{\cot^{2}t}{4\tan t}=-\frac{1}{4}\cot^{3}t$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-\frac{\pi}{4}}=-\frac{1}{4}\cot^{3}(-\frac{\pi}{4})=-\frac{1}{4}(-1)^{3}=\frac{1}{4}$
