Question

I need help verifying a question can someone please help me?

Answer 1

\[(cosx /1+sinx) + 1+sinx/cosx =2secx\]

Answer 2

\[\frac{ cosx }{ 1+sinx } + \frac{ 1+sinx }{ cosx }=2secx\]

Answer 3

okay

Answer 4

yes

Answer 5

i don't see it can you explain it a bit more

Answer 6

but how do I do it with the identities step by step because i have to use the identities to prove it

Answer 7

yea i was looking at this identity sin^2x+cos^2x=1 but if i try changing it around it will look like: sin^2x-1=cos^2x and its supposed to be sin^2x+1 not -1

Answer 8

@zepdrix can you help me please?

Answer 9

@sweetsunray can you help me ?

Answer 10

hey

Answer 11

hi thank you :)

Answer 12

\[\Large\rm \frac{ cosx }{ 1+sinx } + \frac{ 1+sinx }{ cosx }=2secx\]So we've got to show this?

Answer 13

yes

Answer 14

I guess the first thing I'm thinking when I look at this is...
We can make both denominators cos^2x if you put a little work into it.

Answer 15

Err no no, even better.. ummm
Just multiply the first term by the conjugate of the denominator.

Answer 16

\[\Large\rm \frac{ cosx }{ 1+sinx } \color{royalblue}{\left(\frac{1-\sin x}{1-\sin x}\right)}+ \frac{ 1+sinx }{ cosx }=2secx\]

Answer 17

okay ill try it ill tell you what i get

Answer 18

DO NOT distribute out the top, just leave the (1-sin x) in brackets up there.
We're only looking to simplify the denominator right now.

Answer 19

okay

Answer 20

so i got 1-sin^2 x for the denominator

Answer 21

Mmmmk, so going back to your Pythagorean Identity,
can we write that in terms of cosine somehow?

Answer 22

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

Answer 23

yes cos^2x

Answer 24

\[\Large\rm \frac{ \cos x(1-\sin x) }{ \cos^2x } + \frac{ 1+\sin x }{ \cos x }=2\sec x\]Ok great!

Answer 25

See the cancellation we can make from there?
The pieces should start coming together! :)

Answer 26

we can cacel the cosx and the cosx

Answer 27

\[\Large\rm \frac{ 1-\sin x}{ \cos x } + \frac{ 1+\sin x }{ \cos x }=2\sec x\]Good good good.

Answer 28

and now the denominators are the same so you can add the top. 2-sinx /cosx

Answer 29

Woops, your top should be,\[\Large\rm 1+1+\sin x-\sin x\]\[\Large\rm number+number+(potato)-(potato)\]

Answer 30

So we have some potatoes cancelling out, yes?

Answer 31

so its 2/cosx

Answer 32

Yes, and let's go ahead and write it like this,\[\Large\rm 2\left(\frac{1}{\cos x}\right)\]Hmmm, remember any other identities that might help here?

Answer 33

why did you put it like that

Answer 34

I pulled the 2 out of the numerator so we could more easily apply one of our identities.

Answer 35

\[\Large\rm 2\left(\color{orangered}{\frac{1}{\cos x}}\right)\]We have an identity for this orange part.
It wasn't easy to see with the 2 on top though.

Answer 36

ohhh okay because if you multiply its the same as 2/cosx right?
the identity is secx=1/cos and since theres a 2 you multiply so its 2secx

Answer 37

Yayyyy good job!
Not too bad of a problem, right?
Just took a few steps :)

Answer 38

yes thank you for guiding me :) i wish i could do them on my own