Recall the Pythagorean identity,
[tex]\sin^2(\theta) + \cos^2(\theta) = 1[/tex]
Since [tex]\theta[/tex] belongs to Q3, we know both [tex]\sin(\theta)[/tex] and [tex]\cos(\theta)[/tex] are negative. Then
[tex]\cos(\theta) = -\sqrt{1 - \sin^2(\theta)} = -\dfrac7{25}[/tex]
Recall the half-angle identities for sine and cosine,
[tex]\sin^2\left(\dfrac\theta2\right) = \dfrac{1 - \cos(\theta)}2[/tex]
[tex]\cos^2\left(\dfrac\theta2\right) = \dfrac{1 + \cos(\theta)}2[/tex]
Then by definition of tangent,
[tex]\tan^2\left(\dfrac\theta2\right) = \dfrac{\sin^2\left(\frac\theta2\right)}{\cos^2\left(\frac\theta2\right)} = \dfrac{1 - \cos(\theta)}{1 + \cos(\theta)}[/tex]
[tex]\theta[/tex] belonging to Q3 means [tex]180^\circ < \theta < 270^\circ[/tex], or [tex]90^\circ < \theta < 135^\circ[/tex], so that the half-angle belongs to Q2. Then [tex]\sin\left(\frac\theta2\right)[/tex] is positive and [tex]\cos\left(\frac\theta2\right)[/tex] is negative, so [tex]\tan\left(\frac\theta2\right)[/tex] is negative.
It follows that
[tex]\tan\left(\dfrac\theta2\right) = -\sqrt{\dfrac{1 - \cos(\theta)}{1 + \cos(\theta)}} = \boxed{-\dfrac43}[/tex]
