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

1. Loser66

2. Loser66

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

3. anonymous

what identity would i use @Concentrationalizing

4. Loser66

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

5. anonymous

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

6. anonymous

so i would use the Pythagorean identity ? @Concentrationalizing

7. anonymous

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.

8. anonymous

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

9. anonymous

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

10. anonymous

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+

11. anonymous

@Concentrationalizing

12. anonymous

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

13. anonymous

$\frac{ sinx }{ 1-cosx }+\frac{ sinx }{ 1+cosx } = \frac{ sinx(1+cosx)+sinx(1-cosx) }{(1+cosx)(1-cosx) }$ $\frac{ sinx + sinxcosx + sinx - sinxcosx }{ 1-\cos^{2}x }$ $\frac{ 2sinx }{ 1-\cos^{2}x }$ 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 $\frac{ 2sinx }{ \sin^{2}x }= \frac{ 2 }{ sinx } = 2cscx$

14. anonymous

thank you @Concentrationalizing