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Can anyone help me solve a trigonomic identity problem and also help me how to do it step by step?

it's ((cos theta) * (cot theta))/(1-sin theta) - 1 = csc theta

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\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)} \qquad csc(\theta)=\cfrac{1}{sin(\theta)} \\\\\\ sin^2(\theta)+cos^2(\theta)=1\\\\ -------------------------------\\\\


\bf \cfrac{cos(\theta )cot(\theta )}{1-sin(\theta )}-1=csc(\theta )\\\\ -------------------------------\\\\ \cfrac{cos(\theta )\cdot (cos(\theta ))/(sin(\theta ))}{1-sin(\theta )}-1\implies \cfrac{(cos^2(\theta ))/(sin(\theta ))}{(1-sin(\theta ))/(1)}-1\implies \cfrac{cos^2(\theta )}{sin(\theta )}\cdot \cfrac{1}{1-sin(\theta )}-1 \\\\\\ \cfrac{cos^2(\theta )}{sin(\theta )[1-sin(\theta )]}-1\implies \cfrac{cos^2(\theta )-1[sin(\theta )[1-sin(\theta )]]}{sin(\theta )[1-sin(\theta )]}


\bf \cfrac{cos^2(\theta )-1[sin(\theta )-sin^2(\theta )]}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{cos^2(\theta )-sin(\theta )+sin^2(\theta )}{sin(\theta )[1-sin(\theta )]} \\\\\\ \cfrac{cos^2(\theta )+sin^2(\theta )-sin(\theta )}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{\underline{1-sin(\theta )}}{sin(\theta )\underline{[1-sin(\theta )]}} \\\\\\ \cfrac{1}{sin(\theta )}\implies csc(\theta )
User Vinay Hegde
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