Question

Hello. Can someone answe this question? It is a "verify the identity" question. Thank you.

Answer 1

The given identity is:

`\frac{\sin^2\theta\text{ - 1}}{\tan\theta\sin\theta-\tan\theta}=\frac{\sin \theta\text{ + 1}}{\tan \theta\text{ }}`

Let us prove from the Right Hand Side:

`\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}`

The expression becomes:

`\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}`