Question 814201
Eliminate theta from equations x = {{{cosec(theta)}}} and y = {{{expr(1/4)cot(theta)}}}
<pre>
Draw a right triangle and mark an angle <font face="symbol">q</font>

{{{drawing(850/11,100,-.2,3.2,-1.3,4.2,rectangle(2.6,0,3,.4),
locate(.6,.7,theta),   
triangle(0,0,3,0,3,4)  )}}}

We look at the first equation

x = {{{cosec(theta)}}}

Since the cosecant is the hypotenuse over the opposite, and
since x can be written as {{{x/1}}} and x = {{{cosec(theta)}}},
we can put x as the length of the hypotenuse and 1 as the 
length of the opposite.

{{{drawing(850/11,100,-.2,3.2,-1.3,4.2,rectangle(2.6,0,3,.4),
locate(.6,.7,theta), locate(3.1,2,1), locate(.77,2.5,x),  
triangle(0,0,3,0,3,4)  )}}}

And by the Pythagorean theorem we can get the adjacent side

     c² = a² + b²
     x² = a² + 1²
     x² = a² + 1
 x² - 1 = a²

&#8730;<span style="text-decoration: overline">x² - 1</span> = &#8730;<span style="text-decoration: overline">a²</span>

&#8730;<span style="text-decoration: overline">x² - 1</span> = a

We label the length of the adjacent side

{{{drawing(850/11,100,-.2,3.2,-1.5,4.2,rectangle(2.6,0,3,.4),
locate(.6,.7,theta), locate(3.1,2,1), locate(.77,2.5,x),  
triangle(0,0,3,0,3,4),locate(.3,-.2,sqrt(x^2-1))  )}}}

Now we look at the second equation

y = {{{expr(1/4)cot(theta)}}}

The cotangent is the adjacent over the opposite, so we substitute
{{{expr(sqrt(x^2-1)/1))}}} for cot(<font face="symbol">q</font>) or just {{{sqrt(x^2-1)}}}.

in the second equation:

y = {{{expr(1/4)sqrt(x^2-1)}}}

Edwin</pre>