Question 889157
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\cot A\ -\ \tan A}{\csc A\sec A}\ =\ \frac{\frac{\cos A}{\sin A}\ -\ \frac{\sin A}{\cos A}}{\csc A\sec A}\ =\ \frac{\frac{\cos^2 A\ -\ \sin^2 A}{\sin A\cos A}}{\csc A\sec A}\ =\ \frac{\left(\cos^2 A\ -\ \sin^2 A\right)\left(\csc A\sec A\right)}{\csc A\sec A}\ =\ \cos^2 A\ -\ \sin^2 A\ =\ \cos 2A]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos 2A \equiv\ \frac{\cot A\ -\ \tan A}{\csc A\sec A}]


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