1/[(cotX)^2-1]
[(sin^x)^2-(cosx)^2]/[(sinx)^2-sinxcosx]
I've spent the last 3 hours trying to simplify these, and havn't had any luck.
The denominator should remind you of $\displaystyle \sin^2 x + \cos^2 x = 1$, or $\displaystyle \sin^2 x = 1 - \cos^2 x$. Divide both sides of this equation by $\displaystyle \sin^2 x$ and remember the definitions of the complementary trigonometric functions. Most applications of identities will be like this. You can usually manipulate the closest identity into an applicable form.
This has very little to do with trigonometry. It is algebraic factoring. Look at the bigger picture:[(sin^x)^2-(cosx)^2]/[(sinx)^2-sinxcosx]
$\displaystyle \frac{x^2 - y^2}{x^2 - xy} = \cdots$
Can you simplify this fraction?
Try to find $\displaystyle \cot^2 x$ in trigonometric identities you already know. For example, we know the common identity $\displaystyle \sin^2 x + \cos^2 x = 1$. We also know that $\displaystyle \cot x = \frac{\cos x}{\sin x}$. We can thus get an identity involving $\displaystyle \cot^2 x$ by dividing both sides of $\displaystyle \sin^2 x + \cos^2 x = 1$ by $\displaystyle \sin^2 x$. We can then use this identity to simplify our expression.