Since the terminus of [tex]\theta[/tex] lies in quadrant 3, we should have [tex]\cos\theta<0[/tex] and [tex]\sin\theta<0[/tex]. This eliminates the fourth option.
Recall that
[tex]\tan^2\theta+1=\sec^2\theta\implies\sec\theta=\pm\sqrt{\tan^2\theta+1}[/tex]
but we know that [tex]\cos\theta<0[/tex], which means [tex]\sec\theta<0[/tex], so we take the negative root above. Then
[tex]\sec\theta=-\sqrt{\left(\dfrac34\right)^2+1}=-\dfrac54\implies\cos\theta=-\dfrac45[/tex]
so the first option is correct.
We also know that
[tex]\cos^2\theta+\sin^2\theta=1\implies\sin\theta=\pm\sqrt{1-\cos^2\theta}[/tex]
and again we take the negative root here and find that
[tex]\sin\theta=-\sqrt{1-\left(-\dfrac45\right)^2}=-\dfrac35\implies\csc\theta=-\dfrac53[/tex]
so the second option is also correct.
Finally, we have
[tex]\cot\theta=\dfrac1{\tan\theta}=\dfrac43[/tex]
so the third option is also correct.
