Question 713804
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The two sides are NOT identical, in fact they are NEVER equal.


Attempt to solve the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin(A)\ -\ \sin^2(A)}{\cos^2(A)}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A)\ -\ \sin^2(A)\ =\ \cos^2(A)]


Using the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A)\ -\ \sin^2(A)\ =\ 1\ -\ \sin^2(A)]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A)\ =\ 1]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \sin^{-1}(1)\ =\ \frac{\pi}{2}]


But *[tex \LARGE \frac{\pi}{2}] is excluded from the domain of the original equation.  Therefore, there are no real number solutions to the given equation and given that infinite number of counterexamples, the original equation is not an identity.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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