Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

Answer 1

\[(\sin x/1 - \cos x) + (\sin x/1 + \cos x) = 2 \csc x\]

Answer 2

@AnswerMyQuestions help?

Answer 3

@CrazyMexican816 help?

Answer 4

@ilovebmth1234

Answer 5

@mathmath333

Answer 6

@NerdyBookWorm

Answer 7

ok,

Answer 8

sorry im trying to figure this out on paper lol one sec

Answer 9

ok, d\[sinx/1-cosx + sinx/1+cosx = 2cosx\]oes your equation looks similar to this?

Answer 10

2cscx *

Answer 11

Yeah :)

Answer 12

ok well @iGreen

Answer 13

im kinda confused

Answer 14

Any help?

Answer 15

You there?

Answer 16

Can you make the denominators equal? Or Do you know how to do that?

Answer 17

I have no idea how to do that

Answer 18

Okay, It is a method called LCM : Least Common Multiple..

Answer 19

But, I will not go deep into that, because you may confuse, I have one more simpler idea..

Answer 20

\[\frac{\sin(x)}{1 - \cos(x)} \times \frac{1 + \cos(x)}{1 + \cos(x)} + \frac{\sin(x)}{1 + \cos(x)} \times \frac{1 - \cos(x)}{1 - \cos(x)}\]

Can I do like this?

Answer 21

Yeah I assume so

Answer 22

I have multiplied and divided by the same quantity, so it will be cancelled, giving out the same result as we had earlier.. :)

Answer 23

So, I also assume that.. :P

Answer 24

You are not supposed to assume, just look at it and find out why I have written that only..
In other words, in your mind it should strike that this is also a "Rationalization Process".. :)

Answer 25

There, as denominators are same, you can just add the numerator terms.. :)

Answer 26

\[\frac{\sin(x) - \sin(x)\cos(x) + \sin(x) + \sin(x)\cos(x)}{(1-\cos(x)(1+\cos(x))} \implies \frac{2\sin(x)}{1 - \cos^2(x)} \implies \frac{2\sin(x)}{\sin^2(x)}\]

Answer 27

Next step I leave it on you.. :P

Answer 28

I'm so lost. I get what you're saying about the denominators being the same. But I really don't get this.

Answer 29

May I ask something?

You have two persons in your pic, which one is you??

Answer 30

The girl on the left

Answer 31

Type the thing where you have lost.. :)

Answer 32

I was thinking the same, as you have said, I am lost.. :P

Answer 33

See: Look here:

Suppose you have fraction like 5/3.. Now if I multiply it and divide it by same number let us say 4, then our fraction remains same:

\(\frac{5}{3} \times \frac{4}{4}\) = \(\frac{5}{3}\) This you are getting?

Answer 34

Yeah I get that

Answer 35

Look above carefully, I have done the same there..

Answer 36

I have multiplied and divide by the same quantity.. Yes??

Answer 37

\(\large\tt \color{black}{\dfrac{sinx}{1-cosx}+\dfrac{sinx}{1+cosx}}\)

\(\large\tt \color{black}{=sinx(\dfrac{1}{1-cosx}+\dfrac{1}{1+cosx})}\)

\(\large\tt \color{black}{=sinx(\dfrac{1+cosx+1-cosx}{1-cos^2x})}\)

\(\large\tt \color{black}{=sinx(\dfrac{2}{sin^2x})}\)

\(\large\tt \color{black}{=(\dfrac{2}{sinx})}\)

\(\large\tt \color{black}{=2cscx}\)

Answer 38

@mathmath333 again it comes to making denominators equal, and that is the part she is not getting.. :P

Answer 39

@mathmath333 you're a lifesaver. That makes so much sense. Okay thanks :)

Answer 40

Okay my mistake, it was not the denominator part, it was the whole solution which was making her lost.. :P

Answer 41

well i did the same thing as waterinyes
i just took the denominator out