In Exercises 24-34 sketch the parametric curve by eliminating the parameter.

$x=\csc t,y=\cot t,t\in (0,\pi )$

The parametric curve is $x=\csc t,y=\cot t,t\in (0,\pi )$

Consider the parametric equations $x=\csc t,y=\cot t,t\in (0,\pi )$.

The objective is to sketch the parametric curve by eliminating the parameters.

Take the equation $x=csct$.

Square the equation on both sides.

$x^2={\csc }^{2}t$

Take the equation $y=\cot t$.

Square the equation on both sides. Then,

$y^2={\cot }^{2}t$

Now subtract the equations $y^2={\cot }^{2}t$from$x^2={\csc }^{2}t$.

Thus,

$$\begin{gathered} x^2-y^2={\csc }^{2}t-{\cot }^{2}t \\ {x}^2-y^2=1\left[since{\csc }^{2}t-{\cot }^{2}t=1\right] \end{gathered}$$

In order to draw the graph of the equation assume $x=-1,1,2$

Substitute $x=-1$in the equation $x^2-y^2=1.$

Then,

Substitute $x=1$in the equation $x^2-y^2=1$.

Then

$$\begin{gathered} (1{)}^2-y^2-1 \\ 1-y^2=1 \\ y=0\text{simplify} \\ (x,y)=(1,0) \end{gathered}$$

Substitute $x=2$in the equation $x^2-y^2=1$.

Then,

$$\begin{gathered} 2^2-y^2=1 \\ 4-y^2=1 \end{gathered}$$

Add $-4$on both sides of the equation.

$$\begin{gathered} 4-y^2-4=1-4 \\ 4-y^2-4=-3 \\ -y^2=-3 \\ y=\sqrt{3} \\ (x,y)=(2,-\sqrt{3})(2,\sqrt{3}) \end{gathered}$$

The graphical representation by using the points $(1,0)(2,-\sqrt{3})(2,\sqrt{3})$is as follows,

Therefore, the equation after elimination of the parameter is localid="1653400364287" $x^2-y^2=1$