Question 591304
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec(x)\ -\ \cos(x)\ +\ \csc(x)\ -\ \sin(x)\ -\ \left(\sin(x)\tan(x)\right)\ \equiv^?\ \cos(x)\cot(x)]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec(\varphi)\ \equiv\ \frac{1}{\cos(\varphi)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(\varphi)\ \equiv\ \frac{1}{\sin(\varphi)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(\varphi)\ \equiv\ \frac{\sin(\varphi)}{\cos(\varphi)]


to convert everything in the LHS to sine and cosine functions:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos(x)}\ -\ \cos(x)\ +\ \frac{1}{sin(x)}\ -\ \sin(x)\ -\ \frac{\sin^2(x)}{\cos(x)}\ \equiv^?\ \cos(x)\cot(x)]


Combine the first and second and third and fourth terms using *[tex \LARGE \cos(x)] and *[tex \LARGE \sin(x)] as common denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \cos^2(x)}{\cos(x)}\ +\ \frac{1\ -\ \sin^2(x)}{sin(x)}\ -\ \frac{\sin^2(x)}{\cos(x)}\ \equiv^?\ \cos(x)\cot(x)]


Apply the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^2(x)}{\cos(x)}\ +\ \frac{\cos^2(x)}{sin(x)}\ -\ \frac{\sin^2(x)}{\cos(x)}\ \equiv^?\ \cos(x)\cot(x)]


Note the two additive inverse terms in the LHS and eliminate.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\cos^2(x)}{sin(x)}\ \equiv^?\ \cos(x)\cot(x)]


Decompose the fraction in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{\cos(x)}{1}\right)\left(\frac{\cos(x)}{sin(x)}\right)\ \equiv^?\ \cos(x)\cot(x)]


Finally, use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot(\varphi)\ \equiv\ \frac{\cos(\varphi)}{\sin(\varphi)]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\cot(x) \equiv\ \cos(x)\cot(x)]


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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