For a common emitter configuration, if $$\alpha$$ and $$\beta$$ have their usual meanings, the correct relationship between $$\alpha$$ and $$\beta$$ is:

In a transistor we denote the three conventional steady currents by $$I_E$$ for emitter current, $$I_C$$ for collector current and $$I_B$$ for base current. According to Kirchhoff’s current law at the transistor node we have

$$I_E \;=\; I_C \;+\; I_B.$$

Next, by definition of the current gains we write

$$\alpha \;=\;\dfrac{I_C}{I_E} \qquad\text{(common-base current gain)}$$

and

$$\beta \;=\;\dfrac{I_C}{I_B} \qquad\text{(common-emitter current gain)}.$$

From the definition of $$\beta$$ we can express the base current in terms of the collector current:

$$I_B \;=\;\dfrac{I_C}{\beta}.$$

Substituting this expression for $$I_B$$ into the earlier Kirchhoff relation $$I_E = I_C + I_B$$ gives

$$I_E \;=\; I_C \;+\; \dfrac{I_C}{\beta} \;=\; I_C\!\left(1 + \dfrac{1}{\beta}\right).$$

Now we substitute the value of $$I_E$$ in the definition of $$\alpha$$:

$$\alpha \;=\;\dfrac{I_C}{I_E} \;=\;\dfrac{I_C}{I_C\!\left(1 + \dfrac{1}{\beta}\right)} \;=\;\dfrac{1}{1 + \dfrac{1}{\beta}}.$$

To simplify, we multiply numerator and denominator by $$\beta$$:

$$\alpha \;=\;\dfrac{\beta}{\beta\!\left(1 + \dfrac{1}{\beta}\right)} \;=\;\dfrac{\beta}{\beta + 1}.$$

Re-ordering the terms in the denominator we finally obtain

$$\alpha \;=\;\dfrac{\beta}{1 + \beta}.$$

This matches Option A. Hence, the correct answer is Option A.