Q. Prove mathematically that the operating point does not depend on beta, in a potential divider bias circuit ?

To determine the operating point, consider the input section of the voltage divider biasing circuit.

The base current IB is very small compared to the currents in R₁ and R₂.

I₁ ~ I₂

So,

V_B=R₂/ (R₁+R₂) * V_CC

The voltage across the emitter resistor RE

V_E=V₂-V_BE

Then the emitter current is

I_E= V₂-V_BE

And

V_C=V_CC-I_CR_C

V_CE=V_C-V_E = (V_CC-I_CR_C)-I_ER_E

Since collector current is approximately equal to emitter current

V_CE=V_CC-(R_C+R_E) I_C

Since in the above analysis nowhere does beta appear in any equation, which means that operating point does not depend upon the value of beta.