Prove that

Using $\sin \left(90^{\circ}+\theta\right)=\cos \theta$ and $\sin (-\theta)=\sin \theta, \tan \left(90^{\circ}+\theta\right)=-\cot \theta$

$\operatorname{Sin}\left(180^{\circ}+\theta\right)=-\sin \theta($ III quadrant $\sin x$ is negative $)$

$\frac{\cos \theta}{\sin \left(90^{\circ}+\theta\right)}+\frac{\sin (-\theta)}{\sin \left(180^{\circ}+\theta\right)}-\frac{\tan \left(90^{\circ}+\theta\right)}{\cot \theta}=\frac{\cos \theta}{\cos \theta}+\frac{-\sin \theta}{-\sin \theta}-\frac{-\cot \theta}{\cot \theta}$

$=1+(1)-(-1) \Rightarrow 1+1+1=3$