Recall that
[tex]\cos^2\theta+\sin^2\theta=1\implies\sin\theta=-\sqrt{1-\cos^2\theta}[/tex]
where we take the negative square root because we know [tex]\csc\theta=\dfrac1{\sin\theta}<0[/tex]. Then
[tex]\sin\theta=-\sqrt{1-\left(\dfrac49\right)^2}=-\dfrac{\sqrt{65}}9[/tex]
Then by definition of tangent,
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt{65}}9}{\frac49}=-\dfrac{\sqrt{65}}4[/tex]
