Answer
Tangent line: $\quad y=2x-\sqrt{3}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/6}=-3\sqrt{3}$
Work Step by Step
The slope of the tangent line at $t=\displaystyle \frac{\pi}{6}$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=\pi/6}$
$\left\{\begin{array}{l}
\frac{dx}{dt}=\sec t\tan t\\
\\
\frac{dy}{dt}=\sec^{2}t
\end{array}\right.\quad, $
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\sec^{2}t}{\sec t\tan t}=\frac{\mathrm{l}}{\cos t\tan t}=\frac{1}{\sin t}=\csc t$
$\left. \displaystyle \frac{dy}{dx} \right|_{t=\pi/6}=\csc\frac{\pi}{6}=2$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/6}$,
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\frac{d}{dt}[\csc t]}{\sec t\tan t}=\frac{-\csc t\cot t}{\sec t\tan t}=-\cot^{3}t$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/6}=-(\sqrt{3})^{3}=-3\sqrt{3}$
The point $P$ on the curve corresponding to $t=\displaystyle \frac{\pi}{6}$ is
$\displaystyle \left\{\begin{array}{l}
x=\sec(\frac{\pi}{6})=\frac{2}{\sqrt{3}}\\
\\
y=\tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}
\end{array}\right.\quad, P(\frac{2\sqrt{3}}{3},\frac{\sqrt{3}}{3})$
Use the point-slope equation for the tangent line.
$y-\displaystyle \frac{\sqrt{3}}{3}=2(x-\frac{2\sqrt{3}}{3})$
$y=2x-\displaystyle \frac{4\sqrt{3}}{3}+\frac{\sqrt{3}}{3}$
$y=2x-\sqrt{3}$
