Phase shift in an active RC shifter
With a circuit like this one, there are two options: the brute-force approach or the FACTs. For the first one, you can apply the superposition theorem with the input voltage alternatively biasing \$R_1\$ and \$R_3\$ or use the more mathematical KVL-/KCL-based approached described by Jan in his answer.
The second approach that I favor most of the time consists of determining the time constant of this circuit as described by the fast analytical circuits techniques or FACTs. As described in my book on the subject, the principle is simple: determine the time constant involving \$C_1\$ when the excitation is reduced to 0 V (\$V_{in}\$ is replaced by a short circuit) and when the response is nulled. We can start by setting \$s\$ to 0 (dc gain) and see what gain does this circuit exhibit. The below drawing shows the step for this dc analysis. If you do the maths ok (via superposition for instance), you find a gain of 1 regardless of the resistors values:

You then carry on with turning the source off and determining the resistance &quot;seen&quot; from \$C_1\$ terminals. It's easy in this case as it is \$R_1\$ and you have \$\tau_1=R_1C_1\$. Because with a 1st-order circuit the pole is the inverse of the time constant. The pole position is immediate and equal to \$\omega_p=\frac{1}{R_1C_1}\$. For the zero, either you determine the value of \$R\$ when the output is a null or you determine the gain of the circuit at high frequency e.g. when \$C_1\$ is replaced by a short circuit: \$H^1=-\frac{R_4}{R_3}\$. Applying this result by following the guidelines you have a zero located at \$\omega_z=-\frac{R_3}{R_4R_1C_1}\$ and this is a right-half-plane zero (RHPZ).
Since the pole and the zero are located at the same position, they neutralize each other in magnitude and the overall gain is 1 (0 dB) along the frequency axis. However, because of the zero located in the right half-plane, its phase response lags rather than leading as with a classical LHP zero. As a result, it combines with the naturally-lagging response of the pole and brings the phase from 0° down to -180°:

Because \$R_3\$ and \$R_4\$ are of equal values, the pole and the zero solely depend on \$R_1\$ and \$C_1\$. Considering a negative zero, the phase of the expression is thus classically computed as:

This is an all-pass filter which creates a delay whose value depends on the value of \$R_1\$ and \$C_1\$. If you push the maths a little more, this is a way to create a 1st-order Padé approximant of \$e^{-sT}\$ which is a pure delay.