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I have this piece of code, that outputs expression c.

Clear[a, b, c, t, e, r]
l[vektor_] := Sqrt[Total[vektor^2]];     
r = {a Cos[t], b Sin[t]};
e = {Sqrt[a^2 - b^2], 0};
c = Simplify[((r - e)/l[r - e] + (r + e)/l[r + e]).D[r, t],
 {a > b, b > 0, t > 0}]

I know, that c should be 0. Instead of 0, I get this long expression:

enter image description here

Is there any way to simplify this?

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The result is 0, and we can show it by a smart choice of transformation. expr is what I define as the expression that OP wants to simplify:

expr=Sin[t] (b^2 Cos[t] (1/Sqrt[(Sqrt[a^2-b^2]-a Cos[t])^2+b^2 Sin[t]^2]+1/Sqrt[(Sqrt[a^2-b^2]+a Cos[t])^2+b^2 Sin[t]^2])-a ((-Sqrt[a^2-b^2]+a Cos[t])/Sqrt[(Sqrt[a^2-b^2]-a Cos[t])^2+b^2 Sin[t]^2]+(Sqrt[a^2-b^2]+a Cos[t])/Sqrt[(Sqrt[a^2-b^2]+a Cos[t])^2+b^2 Sin[t]^2]))

Clearly FullSimplify does not help even with OP's assumptions:

FullSimplify[expr, a > b > 0 && t > 0]

$\sin(t) \left(b^2 \cos (t) \left(\frac{1}{\sqrt{\left(\sqrt{(a-b) (a+b)}-a \cos (t)\right)^2+b^2 \sin ^2(t)}}+\frac{1}{\sqrt{\left(\sqrt{(a-b) (a+b)}+a \cos (t)\right)^2+b^2 \sin ^2(t)}}\right)-a \left(\frac{a \cos (t)-\sqrt{(a-b) (a+b)}}{\sqrt{\left(\sqrt{(a-b) (a+b)}-a \cos (t)\right)^2+b^2 \sin ^2(t)}}+\frac{\sqrt{(a-b) (a+b)}+a \cos (t)}{\sqrt{\left(\sqrt{(a-b) (a+b)}+a \cos (t)\right)^2+b^2 \sin ^2(t)}}\right)\right)$

We first use the fact that $a>b>0$ to reparametrize $a$ as $b/\sin(k)$ for $\pi/2>k>0$:

expr2=FullSimplify[expr /. a -> b/Sin[k], b > 0 && Pi/2 > k > 0 && t > 0]

$b \cot (k) \sin (t) (\text{sgn}(\csc (k)-\cos (t) \cot (k))-\text{sgn}(\cos (t) \cot (k)+\csc (k)))$

We observe that the expression simplified significantly. Let us now insert back the parameter $a$:

FullSimplify[expr2 /. k -> ArcSin[b/a], a > b > 0 && t > 0]

0

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  • $\begingroup$ Not sure it matters but @galzoidberg assumes a > b. $\endgroup$ – MikeY May 7 at 3:04
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    $\begingroup$ Thanks @MikeY, I used $a>b>0$ in the computations (this is also why I chose $a=b/\sin(k)$) but wrote incorrectly; edited the post now. $\endgroup$ – Soner May 7 at 12:43
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Not an answer, as I could not figure out why, just to confirm that it should be zero

Clear[a, b, c, t, e, r]
L[vektor_] := Sqrt[Total[vektor^2]];
r = {a Cos[t], b Sin[t]}
e = {Sqrt[a^2 - b^2], 0};
c = ((r - e)/L[r - e] + (r + e)/L[r + e]).D[r, t]

Mathematica graphics

Manipulate[
 Plot[c /. {a -> a0, b -> b0}, {t, -200, 200}],
 {{a0, 1, "a"}, -100, 100, 1},
 {{b0, 1, "b"}, -100, 100, 1},
 TrackedSymbols :> {a0, b0}
 ]

Mathematica graphics

And using Chop shows it is zero for any choice of a,b. Reduce also not able to help. FullSimplify did not help either. I think you got Mathematica stumbled on this one.

the only way I could get it to give zero, it to give it bad assumption

 Simplify[c, Sqrt[a^2 - b^2] < 0]
 (* 0 *)

But the above assumption is not correct, since Sqrt[a^2 - b^2] < 0 means complex number is less than 0. But < does not apply to complex numbers, only to real numbers.

Tried Maple's version of Reduce and Maple says it can be zero. Copied the expression to Maple first.

Mathematica graphics

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You say, you know, that c should be zero. Let Reduce test whether it can be unequal zero.

Reduce[{a > b, b > 0, t > 0, c != 0}, {a, b, t}]

(*   False   *)

Edit Another way to show c == 0

Substitute the a-b-squareroute by d and take the b-solution that is > 0.

sol = Solve[Sqrt[a^2 - b^2] == d, b]

(*   {{b -> -Sqrt[a^2 - d^2]}, {b -> Sqrt[a^2 - d^2]}}   *)

c /. sol[[2]] // FullSimplify[#, {a > d > 0, t > 0}] &

(*   0   *)
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To ascertain the nullity we can follow with

s = Normal[Series[c, {t, 0, 5}]] // Factor

and a null common factor appears

$$ a^2 \left(-\sqrt{-2 a \sqrt{a^2-b^2}+2 a^2-b^2}\right)-a^2 \sqrt{2 a \sqrt{a^2-b^2}+2 a^2-b^2}-a \sqrt{a^2-b^2} \sqrt{-2 a \sqrt{a^2-b^2}+2 a^2-b^2}+a \sqrt{a^2-b^2} \sqrt{2 a \sqrt{a^2-b^2}+2 a^2-b^2}+b^2 \sqrt{-2 a \sqrt{a^2-b^2}+2 a^2-b^2}+b^2 \sqrt{2 a \sqrt{a^2-b^2}+2 a^2-b^2} $$

and then

Reduce[{(D[s,t]/.{t->0}) != 0, a > b > 0}, {a, b}]

(* False *)
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