Prove that

Using $\cos \left(90^{\circ}+\theta\right)=-\sin \theta(I$ quadrant $\cos x$ is positive

$\operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta$

$\tan \left(270^{\circ}-\theta\right)=\tan \left(180^{\circ}+90^{\circ}-\theta\right)=\tan \left(90^{\circ}-\theta\right)=\cot \theta$

(III quadrant tanx is positive)

Similarly $\sin \left(270^{\circ}+\theta\right)=-\cos \theta$ (IV quadrant $\sin x$ is negative

$\cot \left(360^{\circ}-\theta\right)=\cot \theta($ IV quadrant $\cot x$ is negative $)$

$=\frac{\sin \left(180^{\circ}+\theta\right) \cdot \cos \left(90^{\circ}+\theta\right) \cdot \tan \left(270^{\circ}-\theta\right) \cdot \cot \left(360^{\circ}-\theta\right)}{\sin \left(360^{\circ}-\theta\right) \cdot \cos \left(360^{\circ}-\theta\right) \cdot \operatorname{cosec}(-\theta) \cdot \sin \left(270^{\circ}+\theta\right)}$

$=\frac{-\sin \theta \cdot-\sin \theta \cdot \cot \theta \cdot-\cot \theta}{-\sin \theta \cdot \cos \theta \cdot-\operatorname{cosec} \theta \cdot-\cos \theta}$

$=\cot \theta \cdot \tan \theta \cdot \cot \theta \cdot \tan \theta \Rightarrow 1$