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sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x
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Nope. Not doing that with no equation editor.
what do you mean
Too hard to type w/o a working equation editor.
how about just like (sinx/(1-cosx)) + (sinx/1-cosx)) = 2 cscx
Add the fractions of the LHS, they have the same denominator. Then multiply the top and bottom of the LHS by (1+cos x), and then do stuff.
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Btw this is NOT an identity. Both sides are not equal.
thank you!
If this is from the class I think it is, the question actually was (sinx/(1-cosx)) + (sinx/1->+<-cosx)) = 2 cscx. Part of the text of the problem is an image, and when you copy & paste, it gives you the wrong thing...
@Astrophysics do you think you could help me with this?
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