Question 1112838
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I have to assume that you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{\sin(b)}\ =\ \frac{\sin(b)}{\cos(b)-1}\ +\ \frac{\sin(b)}{\cos(b)\ +\ 1}]


because resolving the ambiguities any other way doesn't make any sense.  The problem is that the above equation, while it may have a non-null solution set, is not, under any possible definition of the word, an "identity".  An identity is a statement that is true for all possible values of the variable.


As a counter-example to your assertion that this is an identity let *[tex \Large b\ =\ \frac{\pi}{2}].  Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{\sin(\frac{\pi}{2})}\ =\ \frac{\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})-1}\ +\ \frac{\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})\ +\ 1}] 


Since *[tex \Large \sin(\frac{\pi}{2})\ = 1] and  *[tex \Large \cos(\frac{\pi}{2})\ = 0] the statement becomes


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{1}\ =\ \frac{1}{0-1}\ +\ \frac{1}{0\ +\ 1}]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ =\ 0]


Buzz! Sorry, wrong answer.  Thank you for playing.


You also have a big problem if *[tex \Large b\ =\ \pi] or *[tex \Large b\ =\ 0], since either one of those values gives you a zero denominator in the RHS.


So either you wrote the problem incorrectly or your instructor is giving you a trick question.  However, no one should have ever referred to what you wrote as an "identity".


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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