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anonymous
 one year ago
(sinx / one  cosx )+ (sin x /one  cosx) = 2 csc x
anonymous
 one year ago
(sinx / one  cosx )+ (sin x /one  cosx) = 2 csc x

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1How??? check the problem, please.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1but it is true if the denominator is \(1cos^2(x)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what identity would i use @Concentrationalizing

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@Concentrationalizing please, give me the whole thing. Can't wait to see "the 2 more steps" to get the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, I see what you mean. I wasn't paying attention, I apologize.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i would use the Pythagorean identity ? @Concentrationalizing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I was wrong. Loser is right, the problem is faulty. I didnt even bother to check ahead, I just went through the motions of what steps you might normally do.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i'm confused what are the steps i have to do? @Concentrationalizing @Loser66

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The problem is incorrect. You maybe typed something wrong, but your question isn't true.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As loser said, this is the setup that would be correct http://www.wolframalpha.com/input/?i=sinx%2F%281cos^2%28x%29%29+%2B+sinx%2F%281cos^2%28x%29%29+%3D+2cscx+

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, you typed it wrong in your original question then.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ sinx }{ 1cosx }+\frac{ sinx }{ 1+cosx } = \frac{ sinx(1+cosx)+sinx(1cosx) }{(1+cosx)(1cosx) }\] \[\frac{ sinx + sinxcosx + sinx  sinxcosx }{ 1\cos^{2}x }\] \[\frac{ 2sinx }{ 1\cos^{2}x }\] And yes, pythagorean identity from here. \(sin^{2}x + cos^{2}x = 1\) can be rearranged into \(sin^{2}x = 1cos^{2}x\) Thus we have \[\frac{ 2sinx }{ \sin^{2}x }= \frac{ 2 }{ sinx } = 2cscx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you @Concentrationalizing
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