Question 699733
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec^4\theta\ -\ \sec^2\theta\ \equiv^?\ \tan^4\theta\ +\ \tan^2\theta]


Use equivalent sine and cosine expressions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos^4\theta}\ -\ \frac{1}{\cos^2\theta}\ \equiv^?\ \frac{\sin^4\theta}{\cos^4\theta}\ +\ \frac{\sin^2\theta}{\cos^2\theta}]


Apply LCD to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \cos^2\theta}{\cos^4\theta}\ \equiv^?\ \frac{\sin^4\theta\ +\ \sin^2\theta\cdot\cos^2\theta}{\cos^4\theta}]


Factor out *[tex \LARGE \sin^2\theta] from the numerator in the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \cos^2\theta}{\cos^4\theta}\ \equiv^?\ \frac{\sin^2\theta\left(\sin^2\theta\ +\ \cos^2\theta\right)}{\cos^4\theta}]


Apply the Pythagorean Identity


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \cos^2\theta}{\cos^4\theta}\ \equiv^?\ \frac{\sin^2\theta}{\cos^4\theta}]


Apply the Pythagorean Identity again


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \cos^2\theta}{\cos^4\theta}\ \equiv\ \frac{1\ -\ \cos^2\theta}{\cos^4\theta}]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec^4\theta\ -\ \sec^2\theta\ \equiv\ \tan^4\theta\ +\ \tan^2\theta]


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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