Having trouble verifying a trig identity.

I was trying to get above...but I see that I never finished it, like I thought I did.So I guess it was just a mistake. Sorry about that.So, in summary, the given equation is equal to \tan^{2}t, and the two equations \cot^{2}t and \tan^{2}t are not equal.
  • #1
kieth89
31
0

Homework Statement


[itex]\frac{cos^{2}t+tan^{2}t -1}{sin^{2}t} = tan^{2}t[/itex]

Homework Equations


Here are all the trig identities we know up to this point (the one's that we have learned so far, obviously we derive many others from these when verifying identities).

Pythagorean Identities:
[itex]sin^{2}x + cos^{2}x = 1[/itex]
[itex]tan^{2}x + 1 = sec^{2}x[/itex]
[itex]cot^{2}x + 1 = csc^{2}x[/itex]

Reciprocal Identities:
[itex]\frac{sinx}{cosx} = tanx[/itex]
[itex]\frac{1}{sinx} = cosx[/itex]

Even-Odd Identities:
[itex]sin(-x) = -sinx[/itex]
[itex]cos(-x) = cosx[/itex]
[itex]tan(-x) = -tanx[/itex]


The Attempt at a Solution


I've tried this a couple different ways, but still can't figure it out. I'll only show the 3 attempts that I feel came the closest, as these are rather time consuming to type up (But I do think this site does math typing the best I've seen so far). By the third attempt, I began to wonder if this is even a possible identity.

Also, zooming in on your browser will make some of the fractions more readable. Most browsers use the shortcut "Ctrl +" for this.

The first attempt uses the left side (the side with the fraction)
[itex]\frac{cos^{2}t}{sin^{2}t} + \frac{tan^{2}t}{sin^{2}t} - \frac{1}{sin^{2}t}[/itex]

[itex]cot^{2}t + \frac{tan^{2}t}{1-cos^{2}t} - csc^{2}t[/itex]

[itex]csc^{2}t - 1 - csc^{2}t + \frac{tan^{2}t}{1 - cos^{2}x}[/itex]

[itex]\frac{tan^{2}t}{1 - \frac{sin^{2}t}{tan^{2}t}} -1[/itex]

[itex]\frac{tan^{2}t}{\frac{tan^{2}t}{tan^{2}t} -\frac{sin^{2}t}{tan^{2}t}} - 1[/itex]

[itex]\frac{tan^{4}t}{tan^{2}t - sin^{2}t} - 1[/itex]

[itex]\frac{tan^{4}t - tan^{2}t + sin^{2}t}{tan^{2}t - sin^{2}t}[/itex]
...and this is where I give up with this method, not seeing how this will = [itex]tan^{2}t[/itex]

My second attempt I try the right side, that only has the tan
[itex]sec^{2}t - 1[/itex]

[itex]\frac {1}{cos^{2}t} - sin^{2}t - cos^{2}t[/itex]

[itex]\frac{1}{cos^{2}t} - \frac{(cos^{2}t)(sin^{2}t)}{cos^{2}t} - \frac{cos^{4}t}{cos^{2}t}[/itex]

[itex]\frac{1 - (cos^{2}t)(sin^{2}t) - cos^{4}t}{cos^{2}t}[/itex]

[itex]\frac{-1(cos^{4}t + (cos^{2}t)(sin^{2}t) - cos^{2}t - sin^{2}t}{cos^{2}t}[/itex]

[itex]\frac{-1(cos^{4}t + (cos^{2}t)(sin^{2}t) - cos^{2}t - (tan^{2}t)(cos^{2}t)}{cos^{2}t}[/itex]

[itex]\frac{-1(cos^{2}t)(cos^{2}t + sin^{2}t - 1 - tan^{2}t)}{cos^{2}t}[/itex]

[itex]-cos^{2}t - sin^{2}t + 1 + tan^{2}t)[/itex]
...and this is when I give up. As you can see I was grasping at straws toward the end trying to get it figured out.

Attempt number 3, this one makes me think that this isn't even an identity. Started with the fraction side again

[itex]\frac{cos^{2}t + tan^{2}t - sin^{2}t - cos^{2}t}{sin^{2}t}[/itex]

[itex]\frac{tan^{2}t - sin^{2}t}{sin^{2}t}[/itex]

[itex]\frac{tan^{2}t}{sin^{2}t} - \frac{sin^{2}t}{sin^{2}t}[/itex]

[itex]\frac{tan^{2}t}{(tan^{2}t)(\frac{1}{csc^{2}t})} - 1[/itex]

[itex]\frac{tan^{2}t}{\frac{tan^{2}t}{csc^{2}t})} - 1[/itex]

[itex](tan^{2}t)(\frac{csc^{2}t}{tan^{2}t}) - 1[/itex]

[itex]csc^{2}t - 1[/itex]

[itex]cot^{2}t[/itex]
...And here is where I began to wonder if this is a possible identity.

Anyways, as you can hopefully see, I have tried this many different ways and still can't figure it out. I'm probably missing something simple, but I just don't see it. Any help is greatly appreciated.
 
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  • #2
kieth89 said:

Homework Statement


[itex]\frac{cos^{2}t+tan^{2}t -1}{sin^{2}t} = tan^{2}t[/itex]

Homework Equations


Here are all the trig identities we know up to this point (the one's that we have learned so far, obviously we derive many others from these when verifying identities).

Pythagorean Identities:
[itex]sin^{2}x + cos^{2}x = 1[/itex]
[itex]tan^{2}x + 1 = sec^{2}x[/itex]
[itex]cot^{2}x + 1 = csc^{2}x[/itex]

Reciprocal Identities:
[itex]\frac{sinx}{cosx} = tanx[/itex]
[itex]\frac{1}{sinx} = cosx[/itex]

Even-Odd Identities:
[itex]sin(-x) = -sinx[/itex]
[itex]cos(-x) = cosx[/itex]
[itex]tan(-x) = -tanx[/itex]

The Attempt at a Solution


I've tried this a couple different ways, but still can't figure it out. I'll only show the 3 attempts that I feel came the closest, as these are rather time consuming to type up (But I do think this site does math typing the best I've seen so far). By the third attempt, I began to wonder if this is even a possible identity.

Also, zooming in on your browser will make some of the fractions more readable. Most browsers use the shortcut "Ctrl +" for this.

The first attempt uses the left side (the side with the fraction)
[itex]\frac{cos^{2}t}{sin^{2}t} + \frac{tan^{2}t}{sin^{2}t} - \frac{1}{sin^{2}t}[/itex]

[itex]cot^{2}t + \frac{tan^{2}t}{1-cos^{2}t} - csc^{2}t[/itex]

[itex]csc^{2}t - 1 - csc^{2}t + \frac{tan^{2}t}{1 - cos^{2}x}[/itex]
...
Hello kieth89. Welcome to PF !

Rather than changing sin2(t) to 1-cos2(t),

change tan2(t) to [itex]\displaystyle \frac{\sin^2(t)}{\cos^2(t)}[/itex]

Then you're almost there !
 
  • #3
SammyS said:
Hello kieth89. Welcome to PF !

Rather than changing sin2(t) to 1-cos2(t),

change tan2(t) to [itex]\displaystyle \frac{\sin^2(t)}{\cos^2(t)}[/itex]

Then you're almost there !

Aha! I see it now. Thank you very much, that problem really had me stuck.

Also, why was I able, on attempt 3, to get [itex]cot^{2}t = tan^{2}t[/itex]? I'm new to trig stuff, and when I see something like that I think right away not possible (like 3 = 5 and stuff). Of course, then I check it by plugging the initial equations into the calculator and see that they do (approximately) equal each other. But why can those two be equal? I know that cot is just the tangent ratio flipped...I don't know, it just seems weird when I'm able to get that. I very well could have made an error in my calculations, and will look them over after I post this. Just seems weird.
 
  • #4
kieth89 said:
Aha! I see it now. Thank you very much, that problem really had me stuck.

Also, why was I able, on attempt 3, to get [itex]cot^{2}t = tan^{2}t[/itex]? I'm new to trig stuff, and when I see something like that I think right away not possible (like 3 = 5 and stuff). Of course, then I check it by plugging the initial equations into the calculator and see that they do (approximately) equal each other. But why can those two be equal? I know that cot is just the tangent ratio flipped...I don't know, it just seems weird when I'm able to get that. I very well could have made an error in my calculations, and will look them over after I post this. Just seems weird.
[itex]\displaystyle\sin^2(t)\ne \frac{tan^2(t)}{\csc^2(t)}[/itex]

[itex]\displaystyle\sin^2(t)=tan^2(t)\cdot\cos^2(t) =\frac{tan^2(t)}{\sec^2(t)}[/itex]
 
  • #5
SammyS said:
[itex]\displaystyle\sin^2(t)\ne \frac{tan^2(t)}{\csc^2(t)}[/itex]

[itex]\displaystyle\sin^2(t)=tan^2(t)\cdot\cos^2(t) =\frac{tan^2(t)}{\sec^2(t)}[/itex]

Thought it would be something simple like that. Got too careless after the first couple tries. Anyways, thank you for all the help.
 

1. What is a trig identity?

A trig identity is an equation that is always true for any value of the variables involved. It involves trigonometric functions such as sine, cosine, and tangent.

2. Why is verifying a trig identity important?

Verifying a trig identity is important because it allows us to simplify and manipulate complex trigonometric expressions, making them easier to work with in equations and problem solving.

3. How do I know when to use a specific trig identity?

Knowing when to use a specific trig identity comes with practice and familiarity with the different identities. Generally, it is helpful to look for similarities between the given expression and the identities you are familiar with.

4. What are some common techniques for verifying a trig identity?

Some common techniques for verifying a trig identity include using algebraic manipulation, applying Pythagorean identities, and using reciprocal and quotient identities.

5. What should I do if I am having trouble verifying a trig identity?

If you are having trouble verifying a trig identity, it is helpful to review the basic trigonometric identities and practice working with them. You can also break down the expression into smaller parts and work on verifying each part separately.

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