(cot²α - 4) / (cot²α - cotα - 6); Let x = cotα
Therefore (cot²α - 4) / (cot²α - cotα - 6) = (x²- 4) / (x² - x - 6)
(x²- 4) = x² - 2² = (x-2)(x+2) . Difference of two squares.
(x² - x - 6); This is a quadratic expression,
Multiply the first and last terms = -6x²
we think of two expression that multiply to give -6x² and add up to give -x (Middle term)
Those expression are 2x and -3x
(x² - x - 6) = (x² +2x-3x - 6) = x(x+2) -3(x+2) = (x+2)(x-3)
Recall (x²- 4) / (x² - x - 6) = ((x-2)(x+2)) / ((x+2)(x-3)) = (x-2)/(x-3). Cancelling out.
Recall x = cotα, therefore:
(x-2)/(x-3) = (cotα-2)/(cotα-3)
(cot²α - 4) / (cot²α - cotα - 6) = (cotα-2)/(cotα-3)
