The value of
$\frac{\cos \theta }{\sin ({90}^0+\theta )}+\frac{\sin \theta }{\sin ({180}^0+\theta )}+\frac{\cot ({90}^0-\theta )}{\tan \theta }$ is

Prove the following

1.$(1-{\sin }^{2}A)se{c}^{2}A=1$

2.${\sec }^4\theta -{\sec }^2\theta ={\tan }^4\theta +{\tan }^2$

3.$(\sec \theta -\tan \theta {)}^2=\frac{1-\sin \theta }{1+\sin \theta }$

4.$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +}=\frac{1+\sin \theta }{\cos \theta }$