Question

I REALLY NEED THIS ONE IT'S MY LAST ONE.

Verify the identity. Justify each step.



please help and show your work.

Answer 1

`\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)} \qquad % cosecant csc(\theta)=\cfrac{1}{sin(\theta)} \qquad % secant sec(\theta)=\cfrac{1}{cos(\theta)}\\\\ -------------------------------\\\\ \cfrac{sec(\theta )}{csc(\theta )-cot(\theta )}-\cfrac{sec(\theta )}{csc(\theta )+cot(\theta )}=2csc(\theta )\\\\ -------------------------------\\\\`


`\bf \cfrac{(1)/(cos(\theta ))}{(1)/(sin(\theta ))-(cos(\theta ))/(sin(\theta ))}-\cfrac{(1)/(cos(\theta ))}{(1)/(sin(\theta ))+(cos(\theta ))/(sin(\theta ))}\implies \cfrac{(1)/(cos(\theta ))}{(1-cos(\theta ))/(sin(\theta ))}-\cfrac{(1)/(cos(\theta ))}{(1+cos(\theta ))/(sin(\theta ))} \\\\\\ \cfrac{1}{cos(\theta )}\cdot \cfrac{sin(\theta )}{1-cos(\theta )}-\cfrac{1}{cos(\theta )}\cdot \cfrac{sin(\theta )}{1+cos(\theta )}`


`\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}`


`\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\\\\ -------------------------------\\\\`


`\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 )`