The most general spherically symmetric line element is of the form $$\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+2C(r,t)\mathrm{d}t\mathrm{d}r+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2).$$ I am trying to show that with a transformation $t\rightarrow t+f(r,t)$ for some $f(r,t)$, the line element can be written as $$\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2).$$
Using the transformation for $t$, I obtained
$$\mathrm{d}t'=\left(1+\frac{\partial f}{\partial t}\right)\mathrm{d}t+\frac{\partial f}{\partial r}\mathrm{d}r.$$
Using this equation, I substituted it into the general line element to obtain
\begin{align*} \mathrm{d}s^2=&-A(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-2}\mathrm{d}t'^{2}-A(r,t)\left(\frac{\partial f}{\partial r}\right)^2\left(1+\frac{\partial f}{\partial t}\right)^{-2}\mathrm{d}r^2\nonumber\\ &+2A(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-2}\left(\frac{\partial f}{\partial r}\right)\mathrm{d}t'\mathrm{d}r\nonumber\\ &+2C(r,t)\left(1+\frac{\partial f}{\partial t}\right)^{-1}\mathrm{d}t'\mathrm{d}r-2C(r,t)\left(\frac{\partial f}{\partial r}\right)\left(1+\frac{\partial f}{\partial t}\right)^{-1}\mathrm{d}r^2+B(r,t)\mathrm{d}r^2\nonumber\\ &+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2). \end{align*}
Is the equation that is immediately above correct? How do I proceed to show that the line element can be written as $\mathrm{d}s^2=-A(r,t)\mathrm{d}t^2+B(r,t)\mathrm{d}r^2+r^2(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2)$?
