Question 592883
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{1\ -\ \sin x}\ +\ \frac{1}{1\ +\ \sin x}\ \equiv^?\ 2\sec^2x]


The LCD in the LHS is the product of the two denominators.  Since they are binomial conjugates, the product is the difference of two squares.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ +\ \sin x\ +\ 1\ -\ \sin x}{1\ -\ \sin^2x}\ \equiv^?\ 2\sec^2x]


Apply the Pythagorean Identity to the LHS denominator and simplify the LHS numerator


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{\cos^2x}\ \equiv^?\ 2\sec^2x]


Recall that *[tex \LARGE \frac{1}{\cos\varphi}\ \equiv\ \sec\varphi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sec^2x\ \equiv\ 2\sec^2x]


Q.E.D.


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