If $\tan \theta =\frac{3}{4}$ and $\theta$ is not in the first quadrant then $\frac{\sin \left(\frac{\pi }{2}+\theta \right)-\cot \left(\pi -\theta \right)}{\tan \left(\frac{3\pi }{2}-\theta \right)-\cos \left(\frac{3\pi }{2}+\theta \right)}=$

$\tan \theta =\frac{3}{4}\theta \notin Q_1$ 

$\tan \theta \text{is Positive if }\theta \in Q_3$

$\cos \theta =-\frac{4}{5}$

$\cot \theta =\frac{4}{5}$

$\sin \theta =\frac{-3}{5}$

$\frac{\sin \left(90+\theta \right)-\cot \left(180-\theta \right)}{\tan \left(270-\theta \right)-\cos \left(270+\theta \right)}=\frac{\cos \theta +\cot \theta }{\cot \theta -\sin \theta }=\frac{-\frac{4}{5}+\frac{4}{3}}{\frac{4}{3}+\frac{3}{5}}=\frac{8}{29}$