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}
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}