I have some equation,like this $$ \begin{aligned} &\hat{t}_{g o}=\frac{R}{V_{m}}\left(1+\frac{(n+2) \sigma^{2}+(n+1)^{2}(n+2) \sigma_{f}^{2}}{2(2 n+3)(2 n+5)}-\frac{(n+1) \sigma \sigma_{f}}{2(2 n+3)(2 n+5)}\right)\\ &\dot{R}=\frac{dR}{dt}=-V_{m} \cos \sigma\\ &\sigma=\theta-\lambda\\ &\dot{\lambda}=\frac{d\lambda}{dt}=-\frac{V_{m} \sin \sigma}{R}\\ &{\sigma_f}=\theta_{f}-\lambda\\ &\dot{\theta}_{f}=\frac{d\theta_f}{dt}=0 \end{aligned} $$
To solve this problem,this is my code
Clear["Global`*"]
sigma[t] = theta[t] - lambda[t];
sigmaF[t] = thetaF[t] - lambda[t];
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)));
(D[tgo, t] /. {Derivative[1][r][t] -> (-vm*Cos@(sigma[t])),
Derivative[1][lambda][t] -> -(vm*Sin@(sigma[t]))/(r[t]),
Derivative[1][thetaF][t] -> 0}) //
FullSimplify /. {(lambda[t] - theta[t]) ->
sigmat, (lambda[t] - thetaF[t]) -> sigmaft}
get this
The resulting formula is a little bit bug,like this
To solve this problem, I continue with the substitution,like this
I have compared the result with the original answer, and it feels right, but the operation is very complicated and not intuitive. Is there any good way to keep wolfram's output in the same form as the original answer?