Question 808636
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sec^2(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{1}{\cos^2(x)}]


Then by the chain rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{2\cdot\sin(x)}{\cos^3(x)}\ =\ 2\sec^2(x)\tan(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180/4\ =\ 45], so *[tex \LARGE x\ =\ \frac{\pi}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{\pi}{4}\right)\ =\ \cos\left(\frac{\pi}{4}\right)\ =\ \frac{\sqrt{2}}{2}]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\left(\frac{\pi}{4}\right)\ =\ 1]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec\left(\frac{\pi}{4}\right)\ =\ \sqrt{2}]


You can do your own arithmetic.


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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