analysis of BJT PNP type - why they can use voltage divider?
You are correct that base current will result in deviation from the voltage divider equation. However, if the following condition is satisfied the deviation will be small.

\$ R_1 || R_2 &lt;&lt; (\beta + 1)R_E \$

This can be derived by considering the Thevenin equivalent of the one-port network of the base bias resistors. The terminals of this one-port network are the base of the transistor and ground.

The Thevenin voltage is given by a voltage divider, as this is the voltage with the terminals open-circuited i.e. no base current.

\$ V_{th} = V_{cc} * \frac{R_1}{R_1 + R_2}\$

And the Thevenin resistance is the two resistors in parallel (they are in parallel if you zero the DC source). So the equivalent circuit looks like this







simulate this circuit &ndash; Schematic created using CircuitLab

Assuming the transistor is biased in its linear region we can write a KVL equation that gives (assuming I got my directions right)

\$ V_{th} + V_{be} + I_b(R_1 || R_2 + R_E(\beta + 1)) = V_{cc} \$

Rearranging for the base current gives

\$ I_b = \frac{V_{cc} - V_{th} - V_{be}}{R_1 || R_2 + (\beta + 1)R_E}\$

And the base voltage equals the emitter voltage minus a diode drop. The emitter voltage is

\$ V_E = V_{cc} - (\beta + 1) I_b R_E\$

So \$ V_B = V_{cc} - (\beta + 1) I_b R_E - V_{be}\$

Substituting 

\$ V_B = V_{cc} - (\beta + 1) R_E * \frac{V_{cc} - V_{th} - V_{be}}{R_1 || R_2 + (\beta + 1)R_E} - V_{be}\$

Now if \$ R_1 || R_2 &lt;&lt; (\beta + 1)R_E\$ then,

\$ V_B = V_{cc} - (V_{cc} - V_{th} - V_{be}) - V_{be}\$

Which simplifies to 

\$ V_B = V_{th}\$!

So provided that condition is satisfied the base voltage is the same as if it was a voltage divider. It is almost always a good idea to satisfy this condition as it makes design easier and better as you will have trouble making a real circuit have the bias point you want without it.