should I also include the interval $\pi/4$ and $-\pi/4$

Question

$$x(t) = \sec t$$

$$y(t) = \tan t$$

$$ −π/4 < t < π/4$$

Can anyone tell how to plot the graph for this...

As I got the points as

If $t=-\pi/6$ then $X=2/\sqrt 3$ and $Y=-1/\sqrt 3$

If $t=0 $ then X=$1$ and Y=$0$

If $t=\pi/6 $ then $X=2/\sqrt 3$ and $Y=1/\sqrt 3$

Can anyone please tell me how to plot the graph for this should I also include the interval $\pi/4$ and $-\pi/4$

Answer 1

You need to include even $-\pi/4$ and $\pi/4$

Answer 2

If I am correct, the graph is (in terms of $y$ and $x$) of the equation $$x^2-y^2=1$$

I got it as follows: $$$$ For $t\in(-\pi/4,\pi/4)$,

$$x^2=\sec^2(t)$$ $$y^2=\tan^2(t)$$ Now using $\sec^2 \theta - \tan^2\theta=1$, $$x^2-y^2=1$$ This is the graph of a hyperbola