\dfrac{\cos(90°+\alpha )\cdotp \csc (270°+\alpha )\cdotp \tan (180°-\alpha }{\sec(360°-\alpha )\cdotp \sin (180°+\alpha )\cdotp \cot(90°-\alpha ) }

=\dfrac{(-\sin \alpha )\cdotp (-\sec \alpha ) \cdotp (-\tan \alpha ) }{\sec \alpha \cdotp (-\sin \alpha )\cdotp \tan \alpha }

=1