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