$\bf \cfrac{1+cot^2(\theta )}{1+csc(\theta )}=\cfrac{1}{sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\implies \cfrac{1+\frac{cos^2(\theta )}{sin^2(\theta )}}{1+\frac{1}{sin(\theta )}}\implies \cfrac{~~\frac{sin^2(\theta )+cos^2(\theta )}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}\implies \cfrac{~~\frac{1}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}$

$\bf \cfrac{1}{\underset{sin(\theta )}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}\cdot \cfrac{~~\begin{matrix} sin(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{sin(\theta )+1}\implies \cfrac{1}{sin^2(\theta )+sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\ne \cfrac{1}{sin(\theta )}~\hfill$

5 0

3 years ago

-8 - 3x = 19<br> What is the answer

iogann1982 [59]

First, move like terms on one side. Next, add them and divide. See the attachment for solution.

4 0

3 years ago

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