Question

(sinx / one - cosx )+ (sin x /one - cosx) = 2 csc x

Answer 1

How??? check the problem, please.

Answer 2

but it is true if the denominator is \(1-cos^2(x)\)

Answer 3

what identity would i use @Concentrationalizing

Answer 4

@Concentrationalizing please, give me the whole thing. Can't wait to see "the 2 more steps" to get the answer.

Answer 5

Ah, I see what you mean. I wasn't paying attention, I apologize.

Answer 6

so i would use the Pythagorean identity ? @Concentrationalizing

Answer 7

No, 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.

Answer 8

so i'm confused what are the steps i have to do? @Concentrationalizing @Loser66

Answer 9

The problem is incorrect. You maybe typed something wrong, but your question isn't true.

Answer 10

As loser said, this is the set-up that would be correct
http://www.wolframalpha.com/input/?i=sinx%2F%281-cos^2%28x%29%29+%2B+sinx%2F%281-cos^2%28x%29%29+%3D+2cscx+

Answer 11

@Concentrationalizing

Answer 12

Okay, you typed it wrong in your original question then.

Answer 13

\[\frac{ sinx }{ 1-cosx }+\frac{ sinx }{ 1+cosx } = \frac{ sinx(1+cosx)+sinx(1-cosx) }{(1+cosx)(1-cosx) }\]
\[\frac{ sinx + sinxcosx + sinx - sinxcosx }{ 1-\cos^{2}x }\]
\[\frac{ 2sinx }{ 1-\cos^{2}x }\]
And yes, pythagorean identity from here. \(sin^{2}x + cos^{2}x = 1\) can be rearranged into \(sin^{2}x = 1-cos^{2}x\)

Thus we have

\[\frac{ 2sinx }{ \sin^{2}x }= \frac{ 2 }{ sinx } = 2cscx\]

Answer 14

thank you @Concentrationalizing