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Your attempt to animate the tan function misses an essential ingredient.

Consider the trigonometric circle (a unit circle centered at the origin in Cartesian coordinates) and a ray connecting the origin O = (0,0) to a point P = (cos(t), sin(t)).  Then, the horizontal projection of P is (cos(t),0) and the vertical projection is (0,sin(t)).  This enables us to associate the sine and cosine functions with the diagram's geometry.

Question: Where does the tan function fit in that diagram?

Answer:  Draw a tangent line L to the circle at (1,0).  Extend the ray OP to intersect the line L at some point H. You will see right away that H = (1, tan(t)).  So the tan function is associated with the point H in our diagram.  This is an absolutely essential information that should be conveyed to the students when teaching them the trigonometric circle.  [The function cot also has a similar geometric interpretation; I will leave it to you to figure it out.]

The animation below demonstrates the tan function properly in the context of the trigonometric circle.  Note the vertical tangent line to the unit circle.

AnimationCercleTrigoFonctiTrigo-rr.mw

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Acer has offered an answer by modifying your worksheet as you have requested.  But your worksheet is much more complicated than it needs be.  Here is a clean way of producing your diagram.  You should be able to merge this with the improvements suggested by acer.

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restart;
A := {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101};
select(x -> is(modp(x,3)=1), A);

The answer is

{7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97}

By the way, if you intend A to represent the list of primes up to 101, then A should include 2 as well.  Two is the only even prime.

As I noted earlier, it is a good idea to start with a simple case and build upon it after you gain confidence.  This toy example should be enough to get you started.

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PS: Replacing combine with convert on the last line yields a better looking result:

I don't know how the undocumented goto works.  Here is a  version without goto.  Have a look at the help page for break.

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I see a lot of experiments in your worksheet but I didn't see a clear statement of the objective.  Perhaps this is what you have in mind?

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What you have is not an error, but a bad choice of coordinates.  The intersection of a cone and plane is best described in cylindrical coordinates.  You will get a perfect ellipse that way.

You have the cone z=x^2+y^2, and the plane z = 3 + y/3.  Switch to cylindrical coordinates with x=r*cos(t), y=r*sin(t).   Then the equations of the cone and the plane become z = r^2 and z=3+1/3*r*sin(t).  At the intersection we have r^2 = 2 + 1/3*r*sin(t).  Solve this quadratic for r in terms of t.  Pick either of the two solutions. Call it r(t). Then the intersection curve is the ellipse described as a space curve:

[  r(t)*cos(t), r(t)*sin(t), 3 + 1/3*r(t)*sin(t)] ,    -Pi < t < Pi

Double-check your equation.  Perhaps "y" should be "u"?

If so, your boundary value problem does have a solution.  Here it is.

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I can't see a meaning in what you have written.  Perhaps this is what you really want?

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In your worksheet you have things like d*u/dt.  That says "d" times "u" divided by "dt". That's not how one expresses a derivative in Maple. Considering that, I suggest that you begin with learning about the Maple syntax, solve a few simple ordinary differential equations, and then a few simple partial differential equations, and only then attempt to solve a nonlinear system of PDEs.

For whatever it's worth, I have converted your sketches of equations to proper Maple syntax and solved them. I can't tell how useful will this be to you.

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