Deriving an Element of the Lorentz Group SO(1, 3)

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We know that $SO(1, 3)$ is isomorphic to $SU(2) \otimes SU(2)$: $$SO(1, 3) \cong SU(2) \otimes SU(2)$$

We also know that $$ \left(\frac{1}{2}, \frac{1}{2}\right) = \left(\frac{1}{2}, 0\right) \otimes \left(0, \frac{1}{2}\right) $$ where $(j, j')$ is the $(j, j')$ representation of $SO(1, 3)$.

An element of the $\left(\frac{1}{2}, 0\right)$ representation is given by $$ e^{\frac{i}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\phi} \cdot \vec{\sigma}}; $$

and an element of the $\left(0, \frac{1}{2} \right)$ representation is given by $$ e^{\frac{i}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\phi} \cdot \vec{\sigma}}; $$ where $\vec{\theta}$ is the rotation parameters, $\vec{\phi}$ is the boost parameters and $\vec{\sigma}$ is the celebrated Pauli matrices.

How can we evaluate the following direct product to get an element of $\left(\frac{1}{2}, \frac{1}{2} \right)$:

$$ e^{\frac{i}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\phi} \cdot \vec{\sigma}} \otimes e^{\frac{i}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\phi} \cdot \vec{\sigma}}$$