Question

How would i rewrite this equation so it is not in fraction form


-5/(1-cos(-x))

Answer 1

`\bf -\cfrac{5}{1-cos(-x)}\implies -\cfrac{5}{\underset{\textit{symmetry identity}}{1-cos(x)}}\impliedby \begin{array}{llll} \textit{let's multiply top/bottom}\\ \textit{by the conjugate 1+cos(x)} \end{array} \\\\\\ \cfrac{-5}{1-cos(x)}\cdot \cfrac{1+cos(x)}{1+cos(x)}\implies \cfrac{-5(1+cos(x))}{\underset{\textit{difference of squares}}{[1-cos(x)][1+cos(x)]}} \\\\\\`

`\bf \cfrac{-5[1+cos(x)]}{1^2-cos^2(x)}\implies \cfrac{-5-5cos(x)}{\underset{\textit{pythagorean identity}}{1-cos^2(x)}}\implies \cfrac{-5-5cos(x)}{sin^2(x)} \\\\\\ \cfrac{-5}{sin^2(x)}-\cfrac{5cos(x)}{sin^2(x)}\implies -5\cdot \cfrac{1}{sin^2(x)}-5\cdot \cfrac{1}{sin(x)}\cdot \cfrac{cos(x)}{sin(x)} \\\\\\ -5\cdot csc^2(x)-5\cdot csc(x)\cdot cot(x)\implies \boxed{-5csc^2(x)-5csc(x)cot(x)}`