Mathematics, 05.07.2019 15:30, Cobetate
Ireally need this one it's my last one. verify the identity. justify each step. and show your work.
Answers: 1
![\bf \cfrac{sin(\theta )}{cos(\theta )[1-cos(\theta )]}-\cfrac{sin(\theta )}{cos(\theta )[1+cos(\theta )]} \\\\\\ \cfrac{sin(\theta )[1+cos(\theta )]~-~sin(\theta )[1-cos(\theta )]}{cos(\theta )[1+cos(\theta )][1-cos(\theta )]} \\\\\\ \textit{notice that }[1+cos(\theta )][1-cos(\theta )]\textit{ is a difference of squares}](/tpl/images/0054/5128/74816.png)
![\bf \cfrac{\underline{sin(\theta )}+sin(\theta )cos(\theta )\underline{-sin(\theta )}+sin(\theta )cos(\theta )}{cos(\theta )[1^2-cos^2(\theta )]} \\\\\\ \cfrac{2sin(\theta )cos(\theta )}{cos(\theta )[1^2-cos^2(\theta )]}\\\\ -------------------------------\\\\ \textit{recall that }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\qquad thus\\\\ -------------------------------\\\\](/tpl/images/0054/5128/2f527.png)
![\bf \cfrac{2sin(\theta )cos(\theta )}{cos(\theta )[sin^2(\theta )]}\implies \cfrac{2\underline{sin(\theta )cos(\theta )}}{\underline{cos(\theta )sin(\theta )} sin(\theta )}\implies \cfrac{2}{sin(\theta )} \\\\\\ 2\cdot \cfrac{1}{sin(\theta )}\implies 2\cdot csc(\theta )\implies 2csc(\theta )](/tpl/images/0054/5128/cf072.png)
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