Help Verifying Trigonometric Identity I could really use help, hint or otherwise, in proving a trigonometric identity:
We are only allowed to work on one side of the equation.

$$\dfrac{2\sin^2(x)-5\sin(x)+2}{\sin(x)-2} = 2\sin(x)-1$$


HINT: Factorize the numerator and cancel terms arguing why the terms you are canceling are not zero.
Move your mouse over the gray area for the answer.

$$\dfrac{2 \sin^2(x) - 5 \sin(x) + 2}{\sin(x) - 2} = \dfrac{2 \sin^2(x) - 4 \sin(x) - \sin(x) + 2}{\sin(x) - 2}\\ = \dfrac{2 \sin(x) (\sin(x) - 2) - ( \sin(x) - 2)}{\sin(x) - 2}\\= \dfrac{(\sin(x) - 2) (2 \sin(x) - 1)}{\sin(x) - 2} = 2 \sin(x) - 1$$ We are allowed to cancel $\sin(x) - 2$ since $\sin(x) \neq 2$, $\forall x \in \mathbb{R}$.

