simplify(combine(-sqrt(1-cos(t))*sqrt(cos(t)+1)))  assuming t>=0, t<=Pi;

                                     -sin(t)

For other  t  this is not true.

If the variables in your expression are ordinary letters of the English alphabet, extract them as follows:

Expr:="(x-y)/(x-y)";

S:={seq(Expr[i], i=1..length(Expr))};

map(parse, select(StringTools[IsAlpha], S));

The simple procedure  CoeffsOfLinComb   finds all the coefficients of zero linear combination of several polynomials. If all the coefficients are 0, this means that the polynomials are linearly independent. It is clear that in the case of non-zero coefficients they are not uniquely defined, but only up to a constant factor. The procedure works also with polynomials with symbolic coefficients (in general case).

CoeffsOfLinComb:=proc(L::list(polynom), V::list(symbol))

local n, lincomb, M;

n:=nops(L);

lincomb:=expand(add(C[k]*L[k], k=1..n));

M:={coeffs(lincomb,V)};

solve(M, {seq(C[k], k=1..n)});

end proc: 

Examples of use:

CoeffsOfLinComb([x+y, y+z, x+z], [x,y,z]);   # The polynomials are linearly independent

CoeffsOfLinComb([x+y, y+z, x+z, x+y+z], [x,y,z]);  # The polynomials are linearly dependent

map(t->lhs(t)=eval(rhs(t), C[1]=1), %); 

CoeffsOfLinComb([x, a*x+b*y, a*x-b*y], [x,y]);  # The polynomials with symbolic coefficients

map(t->lhs(t)=eval(rhs(t), C[2]=1),%);

I do not understand why you use the loop. A simple solution without any loops:

restart;

Z:=Matrix(2, symbol=z);

Z[1,1]:=1;

Z;

The command  identify does the inverse transformation:

evalf( ln(3) );

evalf( exp(5) );

identify(%%);

identify(%%);

convert(%%, ln);

I understand the question so that it is not allowed to use any packages, only the core of Maple. Solution for rows (similarly for columns):

M:=<1,  5,  6; 0,  0,  0;  2,  3, 5 >:

r:=0:

for i from 1 to op([1,1], M) do

if convert([seq(M[i,j]<>0, j=1..op([1,2],M))], `or`) then r:=r+1 fi;

od:

r;

                                           2

If I understand the problem, all the pieces are in the form of a circular sector of the same radius. The first step - add them together to get a single sector.

1) The solution of 3 sections: just cut this single sector by 3 sections into 4 equal pieces.

2) The solution of 2 sections: one section goes along the symmetry axis of the sector, the second section is perpendicular to the first one. For reasons of continuity it is always possible.

Addition. With Maple (if the radius and the angle of the sector are known) easy to find at what distance from the center of the circle to do the second cut.

 Addition 2. I downloaded the file and saw that I understood the problem incorrectly. Therefore, my solutions are the solutions of other problem. Tomleslie's solution is right.

The equality  int(diff(f(x), x), x=a..b) = f(b) - f(a)  is not always true. For example, consider the integral  int(1/x, x=-1..2) . 

'continuous' option is needed:

Fun := Diff(f(x), x);

int(Fun, x = 2 .. 7, 'continuous');

Since your function  y->int((y⁴+by²+c)^-1/2, y)   is represented by the indefinite integral, for uniqueness, I wrote it as a definite integral  f:=y->Int((s^4+b*s^2+c)^(-1/2), s=0..y). The domain of the inverse function  y=h(x)  for the function  f(y)=x  coincides with the range of  f  when  y=0..infinity.

The procedure  h  numerically solves the problem:

restart;

h:=proc(x,b,c)

global M;

local f;

if b^2-4*c>=0 then error "Should be b^2-4*c<0" fi;

M:=Int((s^4+b*s^2+c)^(-1/2), s=0..infinity);

f:=y->Int((s^4+b*s^2+c)^(-1/2), s=0..y);

if x<0 or x>evalf(M) then error "x does not belong to the range of the function f" fi;

fsolve(f(y)=x, y=0..infinity);

end proc:

The domain of h is  the range  0 .. M .

The examples of use:

h(0.5,2,3);  

h(1.5,2,3);

evalf(M);

plot('h(x,2,3)', x=0..1, color=red, thickness=2);

The procedure  coefff  returns the coefficient of the monomial  t  in the multivariate polynomial  P (T - the list of variables of the polynomial)

coefff:=proc(P, T, t) 

local L, H, i, k:

L:=[coeffs(P, T, 'h')]: H:=[h]: k:=0:

for i from 1 to nops(H) do

if H[i]=t then k:=L[i] fi:

od:

k;

end proc:

Solution of your problem:

Id:=<3*x^2+2*x*y+x, y-x*y+3, y^2-2*x+4>:

L:=[x^2, x*y, x, y^2, y, 1]:

M:=Matrix(3,6, (i,j)->coefff(Id[i], [x,y], L[j]));

Radical of an integer is (in number theory) the product of the distinct primes which divide this integer.

The procedure for calculation:

Rad:=proc(n::integer)

ifactors(n);

`*`(seq(%[2,i,1], i=1..nops(%[2])));

end proc:

Examples:

Rad(54);

Rad(720);

     6

    30

sigma:=167:

t:=i->exp(-i^2/2/sigma^2):

S0:=l->1+2*sum(t(i), i=l..infinity):

S:=l->`if`(l>=32,evalf(S0(l)), evalf(2*add(t(i), i=l..31))+evalf(S0(32))):

S(1),  S(20),  S(31),  S(32),  S(40);  # Examples

              418.6069218, 380.6953067, 358.9442559, 356.9784188, 341.3372758

Obviously, the problem reduces to the calculation  cos(z)  and  sin(z) , if  z=1/2*arccos(1/3) . Using Euler's formula  exp(I*z)=cos(z)+I*sin(z), we obtain

expr:=expand(sin(-(1/6)*Pi+(1/2)*arccos(1/3)));

z:=1/2*arccos(1/3):

evalc(exp(I*z));  # Euler's formula in Maple

eval(expr,[cos(z)=Re(%),sin(z)=Im(%)]);

The easiest way to draw graphs of numerical solutions of differential equations - is to use  plots:-odeplot  command.

Example with 2 functions:

Sol:=dsolve({diff(y[1](x),x)=sin(y[2](x))+2*cos(y[1](x)), diff(y[2](x),x)=2*cos(y[2](x))+3*sin(y[1](x)), y[1](0)=0, y[2](0)=1}, numeric):

plots:-odeplot(Sol, [[x,y[1](x)], [x,y[2](x)]], x=0..4, color=[red,blue], thickness=2, view=[0..4, 0..4]);

Addition. If you still want to use  plot  command, you can write as in example: 

y[1]:=x->rhs(Sol(x)[2]):

plot(y[1], 0..4);

Your contour - it is an ellipse. To maintain the proportions along the axes Ox and Oy is necessary to take the same ranges. For clarity, the cutting plane  z=160   is also shown:

A := plots:-contourplot3d(4*x^2+9*y^2, x = -8.5 .. 8.5, y = -8.5 .. 8.5, contours = [160], color = red, thickness = 4, linestyle = solid):

B := plot3d(4*x^2+9*y^2, x = -8.5 .. 8.5, y = -8.5 .. 8.5, style = surface, view = -15 .. 190):

C:=plot3d(160, x=-8..8, y=-6..6, transparency=0.5):

plots:-display(A, B, C, axes = normal, orientation = [55, 55], lightmodel = light4);