Question

Verify the identity. cot^2(x)/(cosx+1)=(1-sinx)/sinx

Answer 1

I really stink at identity solving, and I think I need a walk through of this one. I can usually figure them out but this one caught me like a deer in headlights.

Answer 2

Convert everything into sine and cosine

Answer 3

\[
\frac{\frac{\cos^2 x}{\sin^2x}}{\cos x+1}=\frac{1-\sin x}{\sin x}
\]

Answer 4

Right, that's what my trig notes tell me. thank you for your response! Once setting up the identity as such, how do you go about simplifying each side?

Answer 5

Get rid of fractions by multiplying the denominators.

Answer 6

How so? This problem has more fractions than I know what to do with! I'm sorry for seeming a little slow to catch on, but I really appreciate the walk through!

Answer 7

Multiply by \(\cos x+1\)

Answer 8

Alright, cool! So on the left side, we're still left with \[\cos^{2}x/\sin^{2}x\]

Answer 9

Sorry, math is really rough for me without a visual representation. I don't think I'm well-versed enough to apply worded explanations to equations without some sort of figure!

Answer 10

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

Answer 11

At this point you want to multiply both sides by \(\sin^2 x\) now.

Answer 12

Right; that way we can get rid of the denominators on both sides at once, correct?

Answer 13

Yes

Answer 14

Awesome! Are we done once we get rid of the denominators? It's mostly a conceptual issue for me.

Answer 15

Hold on...

Answer 16

Sorry, we can't do it the way I've been saying to do it.
I made a mistake.

Answer 17

We cant start with the equation, we have to start with one side and get to the other side.

Answer 18

We have to start with \[
\frac{\cot ^2x}{\cos x+1}
\]

Answer 19

Oh, okay! How should we approach it, then? And also, feel free to put your work on here. This is all for my own personal notes; I just picked a random problem out of my text book to help me understand.

Answer 20

Alright, I follow so far. Sounds simple enough.

Answer 21

To get rid of the fraction, sort of... we do 'multiplication by the conjugate

Answer 22

Now the conjugate of \(\cos x +1\) is \(\cos x -1\)

Answer 23

So we multiply top and bottom by it.

Answer 24

O
kay. I vaguely recall such a method from class but it looks like I opted out of writing it down. Big mistake on my part, it would seem!

Answer 25

Doing that gives us \[
\frac{\cot^2x(\cos x-1)}{\cos^2x -1}
\]

Answer 26

Now when ever we see \(\cos^2\) or \(\sin ^2\) and a \(1\) near by, we want to try using Pythagorean identity.

Answer 27

I follow you thus far.

Answer 28

\[
1=\sin^2x+\cos^2x \\
1-\sin^2x =\cos^2x \\
-\sin^2x =\cos^2x-1

\]

Answer 29

Alright, I'm familiar with this identity; it's the first one we learn! How do I apply this to our previous fraction?

Answer 30

We change the denominator \[
\frac{\cot^2x(\cos x-1)}{-\sin^2x}
\]

Answer 31

Ohhhh! I see now! -sin^{2}x is interchangeable with cos^{2}x-1!

Answer 32

Now what? Are we finished with the left side of the equation?

Answer 33

Now \[
\cot^2x=\frac{\cos^2x}{\sin^2x}
\]

Answer 34

Right, that's what we had before. What do we do with it?

Answer 35

It's on top of -sin^{2}x, right?

Answer 36

Yes.

Answer 37

Alright, just making sure I'm with you - and I am! Go on, please.

Answer 38

I'm wondering it this identity is even true.

Answer 39

Are you sure you wrote it down right?

Answer 40

It's an oddly written identity, that's for sure. I'll see if I can find it in my online textbook. One moment please.

Answer 41

Here it is.

Answer 42

It's \(\csc x\) not \(\cos x\).
Big difference.

Answer 43

Oh no, I'm sorry!!!!!!!!! It's been a long day, I feel awful!

Answer 44

\[
\csc x = \frac 1{\sin x}
\]

Answer 45

Okay, I follow

Answer 46

Sorry, I must go.