Question 584655
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin(\varphi)}{1\ -\ \cos(\varphi)}\ +\frac{1\ -\ \cos(\varphi)}{\sin(\varphi)}\ =^?\ 2\csc(\varphi)]


Apply the LCD and add the two fractions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^2(\varphi)\ +\ 1\ -\ 2\cos(\varphi)\ +\ \cos^2(\varphi)}{\sin(\varphi)\left(1\ -\ \cos(\varphi)\right)}\ =^?\ 2\csc(\varphi)]


Apply the Pythagorean Identity to the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\ -\ 2\cos(\varphi)}{\sin(\varphi)\left(1\ -\ \cos(\varphi)\right)}\ =^?\ 2\csc(\varphi)]


Factor a 2 out of the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\left(1\ -\ \cos(\varphi)\right)}{\sin(\varphi)\left(1\ -\ \cos(\varphi)\right)}\ =^?\ 2\csc(\varphi)]


Eliminate the factor common to both numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{\sin(\varphi)}\ =^?\ 2\csc(\varphi)]


Sine is the reciprocal of Cosecant, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\csc(\varphi)\ \equiv\ 2\csc(\varphi)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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