Trig Proof Question

[Isaiah]

I just cannot figure this one out, but I am sure one of you guys can.

I need to show proof for this trig equation:


(tan^2 x+1)(cos^2 x +1) = tan^2 x +2 (EDIT: Does that look better)?

I tried everyway I can think of... and it only works out if I change a few of the rules

 

[Heisenberg]

Dear god. That's just silly.

 

[CrazyPerson]

do your own h.w

 

[Toastedlightly]

Originally posted by: CrazyPerson
do your own h.w

 

[BigJ]

Are you sure you're writing that out correctly?

EDIT: There we go, I didn't know what the fvck was what.

 

[Rhin0]

Isn't it obviously that .999=1?
That is your answer.
Carry on.

 

[imported_Hi]

Originally posted by: Rhin0
Isn't it obviously that .999=1?
That is your answer.
Carry on.

LOL

 

[Isaiah]

Originally posted by: Rhin0
Isn't it obviously that .999=1?
That is your answer.
Carry on.

Haha... I should I been able to figure that out ;p

 

[ohnnyj]

Originally posted by: BigJ
Are you sure you're writing that out correctly?

 

[Isaiah]

Originally posted by: ohnnyj

Originally posted by: BigJ
Are you sure you're writing that out correctly?

Yes I'm pretty sure I do. I've done 100 proofs like this today... but I just can't figure this one out

[(tan^2)x+1][(cos^2)x +1] = (tan^2)x +2

 

[spidey07]

complete the quadratic equation and then substitue.

 

[ohnnyj]

I've got it.

 

[EugeneChien]

hint - Cos^2 x + Sin^2x = 1

 

[ohnnyj]

You need these two facts:

1. tan^2x = (sin^2x/cos^2x)
2. sin^2x + cos^2x = 1

(tan^2x + 1)(cos^2x + 1)

= sin^2x/cos^2x(cos^2x) + tan^2x + cos^2x + 1

= sin^2x + cos^2x + tan^2x + 1

= tan^2x + 2

 

[EugeneChien]

while we are on the subject, can anyone help me with this?

sec x csc x + ( cot x / ( tan x - 1 ) ) = ( tan x / ( 1 - cot x ) ) - 1

 

[Isaiah]

Originally posted by: ohnnyj
You need these two facts:

1. tan^2x = (sin^2x/cos^2x)
2. sin^2x + cos^2x = 1

(tan^2x + 1)(cos^2x + 1)

= sin^2x/cos^2x(cos^2x) + tan^2x + cos^2x + 1

= sin^2x + cos^2x + tan^2x + 1

= tan^2x + 2

Thanks that really helps

 

[Isaiah]

Originally posted by: EugeneChien
while we are on the subject, can anyone help me with this?

sec x csc x + ( cot x / ( tan x - 1 ) ) = ( tan x / ( 1 - cot x ) ) - 1

Try changing everything to sin and cos before you try to solve it. Sometimes that makes it easier to figure out!

 

[ohnnyj]

Originally posted by: EugeneChien
while we are on the subject, can anyone help me with this?

sec x csc x + ( cot x / ( tan x - 1 ) ) = ( tan x / ( 1 - cot x ) ) - 1

Well dang, that took a while but I finally got it. I don't know if I want to post the whole thing, though as it would take a while.

Some hints:

1. You may need to work backwards as in 1 = sin^2x + cos^2x.
2. You will need to substitute secx = 1/cosx, cscx = 1/sinx, tanx = sinx/cosx, and cotx = cosx/sinx.
3. Simplify fractions (i.e. when you have cotx / (tanx -1)) solve the tanx - 1 part first by substituting (as mentioned in 2. above) and then work from there.
4. It may be helfpul to solve the right-hand side of the equation as well just to see how it should come out.

If you are still stumped let me know. I will be back on around 10:30.

 

[ohnnyj]

Hey Eugene, turn on PMs or I can't help you by that means.

I was going to say:

cos^3x = cosx(1-sin^2x)

See where that takes you.

 

[EugeneChien]

thanks, i got it!

 

[Zugzwang152]

ah fond memories of Algebra II Honors, 11th grade... almost 6 years ago.

 

[Jeff7181]

0.99... = 1

 

[ohnnyj]

Originally posted by: EugeneChien
thanks, i got it!

Sweet, glad I could help.