Question

Prove the identity that

\dfrac{\tan \alpha \cdotp \sin \alpha }{\tan \alpha -\sin \alpha }= \dfrac{\tan \alpha + \sin \alpha}{\tan \alpha \cdotp \sin \alpha}

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

Answer

\tan^2\alpha (1-\sin^2)-\sin^2\alpha =0

\tan^2\alpha -\sin^2\alpha =\tan^2 \alpha\sin^2\alpha

Factor the left hand side using difference of squares formula

(\tan \alpha +\sin \alpha )(\tan \alpha -\sin \alpha )= \tan^2\alpha \sin^2\alpha

Re-arrange the terms and the identity is verified

\dfrac{\tan \alpha \cdotp \sin \alpha }{\tan \alpha -\sin \alpha }= \dfrac{\tan \alpha + \sin \alpha}{\tan \alpha \cdotp \sin \alpha}

Steven Zheng posted 3 months ago

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