Question

i suck at trig plz help with this equation

Answer 1

\[1\div cscx+1 + 1\div cscx- 1\]

Answer 2

\[=2\sec ^{2}xsinx\]

Answer 3

\[\large \frac{1}{\csc x+1}+\frac{1}{\csc x - 1}=2\sec^2x \sin x\]Is this what it's suppose to look like? Try to avoid using the division symbol :) It's really tricky to read sometimes.

Answer 4

yes and sorry

Answer 5

Let's mess with the left side, and try to make it match the right side.

We'll first get a common denominator.
So the first fraction needs to be multiplied by the other term, and the second by the first, to give us a common denom.\[\large \left(\frac{\csc x-1}{\csc x - 1}\right)\frac{1}{\csc x+1}+\frac{1}{\csc x - 1}\left(\frac{\csc x+1}{\csc x +1}\right)\]

Answer 6

k got that step!

Answer 7

Let's look at the first term a sec,
See how on the bottom, they are the SAME just with a different sign between them?
Those are called CONJUGATES. When we multiply conjugates we get the Difference of Squares. Familiar with this? :)

Answer 8

\[\large (a-b)(a+b)=a^2-b^2\]

Answer 9

k

Answer 10

When we apply this rule, we get the square of the first minus the square of the second, giving us a common denominator of,\[\large \csc^2x-1\]

Answer 11

We'll combine the 2 fractions, since they have the SAME denominator.\[\large \frac{(\csc x-1)+(\csc x+1)}{\csc^2x-1}\]

Answer 12

k

Answer 13

got that part!

Answer 14

Hmm it looks like we can simplify the top a little bit :O cancelling out the 1's

Answer 15

\[\large \frac{2\csc x}{\csc^2x-1}\]

Answer 16

Hmm from this point, we need to think back to our identities involving squares.
\[\large \cos^2x+\sin^2x=1\]\[\large 1+\tan^2x=\sec^2x\]\[\large 1+\cot^2x=\csc^2x\]

Hmm that third equation involves cosecant, I think that one will allow us to do something fun here.

Answer 17

If we subtract 1 from both sides of that identity, it gives us,\[\large \cot^2x=\csc^2x-1\]We want to plug this in for the denominator of our problem! :o

Answer 18

so the denominator would be cot^2x

Answer 19

Yessss

Answer 20

\[\large \frac{2\csc x}{\cot^2x}\]Hmmm feels like we're getting close :D

Answer 21

From here, let's just go ahead and convert everything to sines and cosines.. I think that might make things a bit easier to work with.

Answer 22

k

Answer 23

\[\huge \frac{2\frac{1}{\sin x}}{\left(\frac{\cos^2x}{\sin^2x}\right)}\]Understand where those identities came from? :)
And do you remember what we do when we're DIVIDING by a fraction? Another way that we can maybe write that? :O

Answer 24

yes

Answer 25

k the major major part i don't understan dis when there is fractions under factions

Answer 26

This might make it more confusing, but it's worth a shot :) lol
\[\huge \frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)} \qquad \rightarrow \qquad \left(\frac{a}{b}\right)\cdot\left(\frac{d}{c}\right)\]
When we are dividing by a fraction, we can rewrite it as multiplication if we FLIP the bottom fraction, as illustrated in this example.

Answer 27

hmm that makes sense

Answer 28

If you get a some practice with fractions, it might make a little more sense :D
But for now, let's just try to apply it, and see if we can make some sense of it.\[\huge \frac{2\frac{1}{\sin x}}{\left(\frac{\cos^2x}{\sin^2x}\right)} \qquad \rightarrow \qquad 2\frac{1}{\sin x}\cdot \left(\frac{\sin^2x}{\cos^2x}\right)\]

Answer 29

k

Answer 30

Hmm it looks like we can cancel out one of the sine terms.

Answer 31

\[\huge 2\frac{\sin^2x}{\sin x}\cdot \frac{1}{\cos^2x}\qquad \rightarrow \qquad 2\sin x\cdot \frac{1}{\cos^2x}\]

Answer 32

Then we just have 1 final identity to recall.

1/cos^2x = .....something.. Hmmmm

Answer 33

csc^2

Answer 34

no sec

Answer 35

Yesss there we go.

Answer 36

I always remember it based on the fact that the S's and C's DO NOT match up.

Sine with Cosecant, Cosine with Secant

Answer 37

k thank you very much!I am working on the other question and have tried like 5 times still not getting the answer can u just show me the method of the other question that i am goin got ask u?>

Answer 38

i did it right but i am messing it up somewhere idk where

Answer 39

Close this thread, open a new one.

So if someone else can help you, they don't have to dig to the bottom of this thread to find your question.

Type @zepdrix somewhere in the comments, it will alert me that you need assistance and I can find your question easier. I'll try to come help if I have time! :D

Answer 40

k