Question 679329
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A little challenging, but once you see the trick...


Working with the LHS only:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec^4x\ -\ \tan^4x\ =\ \frac{1}{\cos^4x}\ -\ \frac{\sin^4x}{\cos^4x}\ =\ ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos^2x}\,\cdot\,\frac{1}{\cos^2x}\ -\ \frac{\sin^2x}{\cos^2x}\,\cdot\,\frac{\sin^2x}{\cos^2x} \ =\ ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos^4x}\left(1\ -\ \left(\sin^2x\right)\left(\sin^2x\right)\right)\ =\ ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos^4x}\left(1\ -\ \left(1\ -\ \cos^2x\right)\left(1\ -\ \cos^2x\right)\right)\ =\ ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos^4x}\left(1\ -\ \left(1\ -\ 2cos^2x\ +\ \cos^4x\right)\right)\ =\ ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos^4x}\left(2cos^2x\ -\ \cos^4x\right)\ =\ ]


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


which is the same as your right side.


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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