The given identity is:
[tex]\frac{\sin^2\theta\text{ - 1}}{\tan\theta\sin\theta-\tan\theta}=\frac{\sin \theta\text{ + 1}}{\tan \theta\text{ }}[/tex]Let us prove from the Right Hand Side:
[tex]\begin{gathered} \frac{\sin \theta\text{ +1}}{\tan \theta\text{ }} \\ \text{Rationalise the expression by multiplying both the numerator and } \\ \text{denominator by sin}\theta\text{ - 1} \end{gathered}[/tex]The expression becomes:
[tex]\begin{gathered} \frac{\sin\theta\text{ + 1}}{\tan\theta}=\text{ }\frac{(\sin \theta\text{ + 1)(}\sin \theta\text{ - 1)}}{(\tan \theta\text{ )(sin}\theta-\text{ 1)}} \\ \frac{\sin\theta\text{ + 1}}{\tan\theta}=\text{ }\frac{\sin ^2\theta-\sin \theta\text{ + sin}\theta\text{ - 1}}{\tan \theta\sin \theta-\tan \theta} \\ \frac{\sin\theta\text{ + 1}}{\tan\theta}\text{ = }\frac{\sin ^2\theta\text{ - 1}}{\tan \theta\sin \theta-\tan \theta}\text{ (Proved)} \end{gathered}[/tex]