Transfer function of an RC circuit
If you use the fast analytical circuits techniques or FACTs, you'll obtain the transfer function by inspecting the circuit without writing a single line of algebra. Furthermore, it will be presented in a low-entropy form giving you insight on where the poles and zeroes are placed.

Start with \$s=0\$: open the capacitor, what is the gain \$H_0\$ linking \$V_{in}\$ (the stimulus) to the response \$V_{out}\$? \$H_0=1\$.

Now, reduced the excitation to 0 V (replace \$V_{in}\$ by a wire) and "look" through the capacitor's connections to determine the resistance: you "see" that \$R_1\$ and \$R_2\$ are in series leading to a time constant \$\tau=(R_1+R_2)C_1\$. The pole in this 1st-order system is the inverse of the time constant: \$\omega_p=\frac{1}{(R_1+R_2)C_1}\$.

For the zero, look for the condition nulling the response: if you apply a stimulus \$V_{in}\$ what condition would make \$V_{out}=0\$? If the series connection of \$C_1\$ and \$R_2\$ forms a short circuit:\$R_2+\frac{1}{sC_1}=0\$. Solving this simple equation gives a zero \$\omega_z=\frac{1}{R_2C_1}\$.

The complete transfer function is therefore: \$H(s)=H_0\frac{1+\frac{1}{\omega_z}}{1+\frac{1}{\omega_p}}\$. This is the low-entropy form you need to adopt to express transfer functions the right way.

Nothing complicated here, just a few steps to determine the time constants in two conditions: when the stimulus is turned off and when the response is nulled.