For a transistor $$\alpha$$ and $$\beta$$ are given as $$\alpha = \frac{I_c}{I_E}$$ and $$\beta = \frac{I_c}{I_B}$$. Then the correct relation between $$\alpha$$ and $$\beta$$ will be:

We start with the two standard current-gain definitions for a transistor. By definition we have

$$\alpha = \frac{I_c}{I_E} \qquad\text{and}\qquad \beta = \frac{I_c}{I_B}.$$

Here $$I_E$$ is the emitter current, $$I_c$$ is the collector current and $$I_B$$ is the base current. The three currents are related by Kirchhoff’s current law at the transistor junction, which gives the identity

$$I_E = I_c + I_B.$$

We now want to eliminate the currents step by step so that only $$\alpha$$ and $$\beta$$ remain.

First, from the definition of $$\alpha$$ we can express the collector current $$I_c$$ in terms of $$I_E$$:

$$I_c = \alpha I_E.$$

Next, substitute this expression for $$I_c$$ in the current‐sum relation $$I_E = I_c + I_B$$. We obtain

$$I_E = \alpha I_E + I_B.$$

To make the equation easier to work with, bring the $$\alpha I_E$$ term to the left side:

$$I_E - \alpha I_E = I_B.$$

Factor out $$I_E$$ on the left:

$$(1-\alpha) I_E = I_B.$$

Now isolate $$I_E$$ because it will be convenient in the next step:

$$I_E = \frac{I_B}{1-\alpha}.$$

Return to the definition of $$\beta$$, namely $$\beta = \dfrac{I_c}{I_B}$$. We already have $$I_c = \alpha I_E$$, so substitute this value of $$I_c$$:

$$\beta = \frac{\alpha I_E}{I_B}.$$

But a moment ago we expressed $$I_E$$ as $$I_E = \dfrac{I_B}{1-\alpha}$$. Substitute this expression for $$I_E$$ as well:

$$\beta = \frac{\alpha \left( \dfrac{I_B}{1-\alpha} \right)}{I_B}.$$

The factor $$I_B$$ appears in both numerator and denominator, so it cancels out completely:

$$\beta = \frac{\alpha}{1-\alpha}.$$

Thus we have derived the required relation between the transistor gains $$\alpha$$ and $$\beta$$:

$$\boxed{\displaystyle \beta = \frac{\alpha}{1-\alpha}}.$$

Among the options provided, this matches option C.

Hence, the correct answer is Option 3.