In Maple, the number pi should be encoded as Pi, not pi.

exp(I*Pi);

   #   -1

In my opinion, this whole check is so obvious and is done manually in a few seconds (using periodicity of sinus)  that it seems a bit strange to involve Maple here. 

@jalal  Use  Matrix(2,2, ... )  instead of  Vector[row]

@SSMB I understood your original question as simply replacing the symbols in the f[3] expression with some expressions you specified. I am completely unfamiliar with the specific issues involved in your task. I can't help you any further.

@mmcdara  The variant with

n->Matrix(n,(i,j)->a[i,j])

is more universal, I often use it and therefore remembered it well. It is suitable when instead of  a[i, j]  stands any expression , depending on  i  and  j .

@jalal 

restart;
F:=proc(t)
local P1, P2;
uses plots;
P1:=plot3d([y^2*cos(alpha),y,y^2*sin(alpha)],y=1..2,alpha=Pi/2..t, color="LightGreen");
P2:=spacecurve([y^2*cos(t),y,y^2*sin(t)],y=1..2, color=red, thickness=3);
display(P1,P2);
end proc: 
P:=plots:-spacecurve([0,t,t^2], t=-3..3, color=red, thickness=3):
P1:=plot([[0,0]], x=-4..4,y=-3..3);
f:=plottools:-transform((x, y)->[x, y, 0]):
plots:-animate(F,[t], t=Pi/2..2*Pi+Pi/2, frames=90, labels=[x,y,z], background=P, paraminfo=false, axes=normal, axis[3]=[color=white],view=[-4..4,-3..3,-4..9]);

@jalal  The following works :

restart;
F := t->plot3d([(y-1)*(y+2)*y*cos(alpha), y, (y-1)*(y+2)*y*sin(alpha)], y = -2.5 .. 0.5, alpha = (1/2)*Pi .. t, color = "LightBlue"):
P := plots:-spacecurve([0, t, (t-1)*(t+2)*t], t = -3 .. 2, color = red, thickness = 2):
plots:-animate(F, [t], t = (1/2)*Pi .. 2*Pi+(1/2)*Pi, frames = 90, labels = [x, y, z], background = P, paraminfo = false, axes = normal, view = [-4 .. 4, -3 .. 3, -4 .. 9], orientation=[-20,65]);

@nm  For problems with a parameter in Mathematica there is a very powerful  Reduce  command that does a full analysis of an equation or inequality:

I think developers should pay attention to this post, because none of the natural approaches give the desired result (code in Maple 2018.2):

solve( x^2+(y-2)^2=0, {x,y}) assuming real;  # Incorrect output
RealDomain:-solve( x^2+(y-2)^2=0, {x,y});  # NULL

@acer Thank you for your efforts in analyzing this situation in detail.

@Carl Love  and  @mmcdara  Thanks for the answers!

We get the following,

is(Or(cos(x) = -sqrt(1-sin(x)^2), cos(x) = sqrt(1-sin(x)^2))) assuming real;

  #  FAIL

It's a shame Maple doesn't know the right answer  true .

@mmcdara  The formula  cos(x)=(-4*sin(x/2)^4+sin(x)^2)/(4*sin(x/2)^4+sin(x)^2)
  may lead to errors for some argument values (i.e. these expressions are not equivalent). In addition, the expression becomes more complex. I still don't understand what the meaning of these transformations is?

eval(cos(x)=(-4*sin(x/2)^4+sin(x)^2)/(4*sin(x/2)^4+sin(x)^2), x=0);

       Error, numeric exception: division by zero

@Andiguys Yes, it is easy to do. Just remove x=10000 from your data. To plot 2 graphs for each vertical axis, use a separate list for each pair of graphs. Read the plot,details help for this.

 vv ,  mmcdara   Thank you for your interest and the solutions provided. My solution below differs from vv's solution in that I use  simplify  with siderals instead of  solve . Since  siderals  does not work with  abs, in order not to miss solutions I had to take into account different signs distributions (4 options in total):

restart;
A:=<x1,a>: B:=<x2,b>: C:=<x3,c>: vA:=<v1,0>: vB:=<v2,0>: vC:=<v3,0>:
f:=t->A+t*vA: g:=t->B+t*vB: h:=t->C+t*vC:
S:=t->1/2*LinearAlgebra:-Determinant(<g(t)-f(t) | h(t)-f(t)>):
P:=combinat:-permute([1,1,-1,-1],2):
{seq(simplify(abs(S(10)), {S(0)=2*p[1], S(5)=3*p[2]}), p=P)};

PS.  This problem is taken from a wonderful book (in Russian)  В.В. Прасолов "Задачи по планиметрии (часть 2)", which contains many original and non-trivial geometric problems on a variety of topics. This book can probably be found in translations into other languages.

If the number (-1) in the exponent simply means the reciprocal value, then the syntax is very simple

simplify(1/(diff(z, t)-diff(z, x)));

If you mean the inverse function, then clarification is required here, since you have a function of several variables.