If [tex]\theta=\tan^{-1}\dfrac43[/tex], then [tex]\tan\theta=\dfrac43[/tex] and so [tex]\cot\theta=\dfrac34[/tex]. Recall the Pythagorean identity,
[tex]\csc^2\theta=\cot^2\theta+1\implies\csc\theta=\pm\sqrt{\cot^2\theta+1}[/tex]
[tex]\theta[/tex] lies in the first quadrant, so we know [tex]\sin\theta>0[/tex], which also means [tex]\csc\theta=\dfrac1{\sin\theta}>0[/tex], so we should take the positive square root. Then
[tex]\csc\theta=\sqrt{\left(\dfrac34\right)^2+1}=\dfrac54[/tex]
