It's exactly the same as the original answer by simplification [closed]

Question

I use this

Clear["Global`*"]
sigma[t] = theta[t] - lambda[t];
sigmaF[t] = thetaF[t] - lambda[t];


rule1 = {Derivative[1][r][t] -> (-vm*Cos@(sigma[t])), 
   Derivative[1][lambda][t] -> -(vm*Sin@(sigma[t]))/(r[t]), 
   Derivative[1][thetaF][t] -> 0};
rule2 = {sigmat -> \[Sigma], sigmaft -> Subscript[\[Sigma], f], 
   vm -> Subscript[V, m], Derivative[1][theta][t] -> 
\!\(\*OverscriptBox[\(\[Theta]\), \(.\)]\)};

tgo = r[t]/
    vm*(1 + ((n + 2)*
         sigma[t]^2 + (n + 1)^2*(n + 2) sigmaF[t]^2)/(2*(2 n + 
          3)*(2 n + 5)) - (n + 1)*sigma[t]*
      sigmaF[t]/(2*(2*n + 3)*(2*n + 5)));

dtgo = ((D[tgo, t] /. rule1) // 
     FullSimplify) /. {(lambda[t] - theta[t]) -> 
     sigmat, (lambda[t] - thetaF[t]) -> sigmaft};
dtgo /. rule2 // 
 Collect[#, {Subscript[V, m], Cos[\[Sigma]], Sin[\[Sigma]], 
\!\(\*OverscriptBox[\(\[Theta]\), \(.\)]\), r[t]}] &

answer is

$$ \begin{aligned} &-\cos \sigma \cdot\left(1+\frac{(n+2) \sigma^{2}+(n+1)^{2}(n+2) \sigma_{f}^{2}}{2(2 n+3)(2 n+5)}-\frac{(n+1) \sigma \sigma_{f}}{2(2 n+3)(2 n+5)}\right) \\ &+\frac{R}{2(2 n+3)(2 n+5) V_{\mathrm{M}}}\left[(2 n+4) \sigma-(n+1) \sigma_{\mathrm{f}}\right] \dot{\theta} \\ &+\frac{1}{2(2 n+3)(2 n+5)}\left[(2 n+3) \sigma+\left(2 n^{2}+6 n+3\right)_{\sigma_{f}}\right] \sin \sigma \end{aligned} $$

use wolfram get this:

By comparing the output with the original answer, you can see some differences

the $\sigma_f$ not collect

There are other problems

and

I've listed some of them here, not all of them, but anyway, I want wolfram's output to be exactly the same as the answer