Answer
$x^{2}-y^{2}=1,\qquad x \lt 0$
.
Work Step by Step
Pythagorean trigonometric identity: $\sec^{2}t=\tan^{2}t+1$
leads to
$x^{2}=y^{2}+1,\qquad x \lt 0$
$x^{2}-y^{2}=1,\qquad x \lt 0$
This is the left wing of a hyperbola.
To graph, create a function value table using values for t, in ascending order of t, calculating the x- and y-coordinates of points on the graph.
Plot and join the points obtained with a smooth curve, noting the direction in which t increases.
