Question

Verify the identity.

\(\sf\dfrac{cos~x}{1+sin~x}~+~\dfrac{1+sin~x}{cos~x}~=~2~sec~x\)

Answer 1

you figured it out ? :)

Answer 2

No

Answer 3

you may start by combining both the fractions on the left hand side

Answer 4

I would have to have a common denominator so, \(\sf\dfrac{(cos~x)(1~+~sin~x)}{(1~+~sin~x)(cos~x)}~+~\dfrac{(1~+~sin~x)(cos~x)}{(cos~x)(1~+~sin~x)}\)

\(\sf\dfrac{cos x + sin x cos x}{cos x + sin x cos x}~+~\dfrac{cos x + sin x cos x}{cos x + sin x cos x}\)

\(\sf\dfrac{2(cos x + sin x cos x)}{cos x + sin x cos x}\)

Answer 5

@e.mccormick

Answer 6

Hmmm.... Your representations of 1 to do common denomintor seem wrong.

\(\sf\dfrac{cos~x}{1+sin~x}~+~\dfrac{1+sin~x}{cos~x}~=~2~sec~x\)

\(\sf\dfrac{cos~x}{1+sin~x}\cdot\dfrac{cos~x}{cos~x}~+~\dfrac{1+sin~x}{cos~x}\cdot\dfrac{1+sin~x}{1+sin~x}~=~2~sec~x\)

Answer 7

They key is that anything divided by itself is 1 and therefore does not change the fraction.

\(\dfrac{1}{4}=\dfrac{1}{4}\cdot\dfrac{a}{a}\)

So to get a cos x onto the bottom of the left most fraction, I multiply by 1 in the form of \(\dfrac{\cos x}{\cos x}\). Similar rules for doing the other fraction to get the common denominator.

Answer 8

Do you see how fixing that mistake will lead to a different answer?

Answer 9

Yes

Answer 10

One of the key things to remember in math is that there are two basic operations that do not change things. First is the adding of 0 and second is multiplying by 1. So why is this useful? Well, let me look at them each:

+42-42 = 0, therefore you can add +42-42 to anything and it will not change the value of an equation. If you need to factor something with a +42, that could make it more workable. You use this basic tactic in completing the square.

Multiplying by 1 I just showed. If it is 1, \(\tfrac{1}{1}\), \(\tfrac{42}{42}\), \(\tfrac{\sin\theta}{\sin\theta}\), or whatever else over itself, it does not matter. It is still multiplying by 1.

You need to keep tat bit of algebra in mind when doing trig. The algebraic manipulations of trig allow you to change the appearance of an equation without changing the actual equality.

Answer 11

Wow

Answer 12

I think I made a mistake while changing the denominator

Answer 13

Yah, which is what my corrected version addressed. When you did it, you did not multiply by 1.

Answer 14

Yes, I accidentally did the first fraction times 1+sinx/cos x and the second one times cos x/1+sin x

Answer 15

When I redo it, I got \(\sf\dfrac{cos^2x}{(1+sin~x)(cos~x)}\) and \(\sf\dfrac{1+sin^2x}{(cos~x)(1+sin~x)}\)

Answer 16

Yep.

Anther thing to keep in mind with this sort of problem is the goal, 2sec x. Well, sec x is 1/cos x. So you know that you need to end up with a cos on the bottom. That means that the 1 + sin x on the bottom must canel out somehow. So leave them factored like that.

OK. So, do the addition from that point and what does it become?

Answer 17

And I think you have a mistake on the right fraction. On the top.

Answer 18

\(\sf\dfrac{1+cos^2x+sin^2x+2~sin~x}{2((1+sin~x)(cos~x))}\)

Answer 19

\((a + b)(a + b)=a^2+2ab+b^2\)

Apply that rule or foil the top of the right and try again. =)

Answer 20

Yah. Tht fixes most of it... but how did you get a 2 on the bottom?

Answer 21

haha yeah, I just added it in on that one, but it should be \(\sf\dfrac{sin^2x+2~sin~x+1}{(1+sin~x)(cos~x)}\)

Answer 22

Whoops, I forgot you don't add the denominators

Answer 23

\(\sf\dfrac{1+cos^2x+sin^2x+2~sin~x}{(1+sin~x)(cos~x)}\)

Answer 24

Yah. OK. So, now you have one fraction, but more still needs to go. Do you see anything else that can be gotten rid of or replaced via an identity?

Answer 25

\(\sf cos^2x+sin^2x=1\)

Answer 26

=) Yep!

Answer 27

So now it would be \(\sf\dfrac{2+2~sin~x}{(1+sin~x)(cos~x)}\)

Answer 28

Yep! Now...see the last thing you need to get rid of?

Answer 29

1+sin x?

Answer 30

Yep. Can you think of what needs to be done to have that on both the top and bottom so that it cancels?

Answer 31

Welp, I need to head out, but at this point you are just a couple of simple steps away from done. =)

Answer 32

:(

Answer 33

Yah, well, there were some service calls. Did you figure out the last steps? It is more algebra than trig at that point.

Trig uses a ton of algebra. Calculus uses a ton of EVERYTHING before it. I mean all the way down to basic addition and subtraction on up to the most complex trig identity swaps. So it is good to keep up on earlier math to make later math easier to do.

Answer 34

No, I wasn't sure how to do the rest :/

Answer 35

Well, you have 2+2sx on the top but need the 1+sx on the bottom to cancel. How can you mathematically make the top look more like the bottom?

Answer 36

\(\sf\dfrac{2+sin~x}{cos~x}\)?

Answer 37

Ummm..... I am not sure what you did there. Think of it as an algebra problem at this point.

\(\dfrac{2+2a}{(1+a)b}\)

It is really the same thing as that. Just did a little substitution. How can I cancel the term (1+a)? What do I need to do to the top to make it work?

Answer 38

\(\sf\dfrac{2(1+sin~x)}{(1+sin~x)(cos~x)}\)

Answer 39

=)

OK. So, now it can cancle. That gets you very close to done.

Answer 40

Ok, so after it cancels it is just \(\sf\dfrac{2}{cos~x}\) right?

Answer 41

Yep! Now, do you see how close that is to \(2\sec x\)?

Answer 42

That would be equal to \(\sf 2sec~x\)

Answer 43

Yah. Cause the 2 pops off the top, leaving \(2\dfrac{1}{\cos x}\)

Answer 44

:)

Answer 45

Yah, well, there were some service calls.

service calls?

Answer 46

I know this one was a bit of a process. You migh want to see if you can make some clean notes on how it worked. That way it will be more likey to make sense if you see something similar at another time.

Answer 47

Thanks :D

Answer 48

And yes, service calls. As in get in a truck, drive places, and see why their internet is not working.

Answer 49

hmmmm sounds interestina

Answer 50

*interesting

Answer 51

Yah, well, you should see the view if it is on a tower.

Answer 52

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