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The correct relation for $\alpha$, $\beta$ for a transistor :
A
$\beta = \dfrac{1 - \alpha}{\alpha}$
B
$\beta = \dfrac{\alpha}{1 - \alpha}$
C
$\alpha = \dfrac{\beta - 1}{\beta}$
D
$\alpha\beta = 1$
Detailed Solution
$\alpha = \dfrac{I_C}{I_E}$ and $\beta = \dfrac{I_C}{I_B}$, with $I_E = I_B + I_C$.
$\beta = \dfrac{I_C}{I_E - I_C} = \dfrac{I_C/I_E}{1 - I_C/I_E}$
$\beta = \dfrac{\alpha}{1 - \alpha}$
(Equivalently $\alpha = \dfrac{\beta}{1 + \beta}$.)
$\beta = \dfrac{I_C}{I_E - I_C} = \dfrac{I_C/I_E}{1 - I_C/I_E}$
$\beta = \dfrac{\alpha}{1 - \alpha}$
(Equivalently $\alpha = \dfrac{\beta}{1 + \beta}$.)