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 ) }



Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 1 week ago


Answer

\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

Steven Zheng Steven Zheng posted 1 week ago

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