![\bf \textit{recall that }sin^2(\theta)+cos^2(\theta)=1\\\\ -------------------------------\\\\ \cfrac{sin(\theta )}{1-cos(\theta )}+\cfrac{1-cos(\theta )}{sin(\theta )}=\cfrac{2}{sin(\theta )}\\\\ -------------------------------\\\\ \cfrac{sin^2(\theta )~~+~~[1-cos(\theta )]^2}{[1-cos(\theta )][sin(\theta )]}\implies \cfrac{sin^2(\theta )~~+~~[1^2-2cos(\theta )+cos^2(\theta )]}{[1-cos(\theta )][sin(\theta )]}](/tpl/images/0249/2559/fd84b.png)
![\bf \cfrac{sin^2(\theta )+cos^2(\theta )~~+~~1-2cos(\theta )}{[1-cos(\theta )][sin(\theta )]}\implies \cfrac{\boxed{1}~~+~~1-2cos(\theta )}{[1-cos(\theta )][sin(\theta )]} \\\\\\ \cfrac{2-2cos(\theta )}{[1-cos(\theta )][sin(\theta )]}\implies \cfrac{2\underline{[1-cos(\theta )]}}{\underline{[1-cos(\theta )]}[sin(\theta )]}\implies \cfrac{2}{sin(\theta )}](/tpl/images/0249/2559/d9723.png)
How do you get it to equal 2/sin(theta)?
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Mathematics, 21.09.2019 11:30 krystinayagel013
How do you get it to equal 2/sin(theta)?
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