Trig Proof Question

Isaiah

Senior member
May 31, 2000
453
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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 ;)
 

BigJ

Lifer
Nov 18, 2001
21,330
1
81
Are you sure you're writing that out correctly?

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

Isaiah

Senior member
May 31, 2000
453
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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
 

ohnnyj

Golden Member
Dec 17, 2004
1,239
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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

Member
Feb 6, 2002
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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

Senior member
May 31, 2000
453
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0
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

Senior member
May 31, 2000
453
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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

Golden Member
Dec 17, 2004
1,239
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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

Golden Member
Dec 17, 2004
1,239
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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.