Thread: A stubborn identity proof.

Hello again, guys. I need to prove the following identity:



After fiddling around with this for an hour and getting nowhere, I decided to bring it here for help. None of the identities I'm using seem to be of much help.

Thanks a bunch!

Originally Posted by JennyFlowers

Hello again, guys. I need to prove the following identity:



After fiddling around with this for an hour and getting nowhere, I decided to bring it here for help. None of the identities I'm using seem to be of much help.

Thanks a bunch!

If you simplify, you will get
[tan^2(θ) - cot^2(θ)]/tanθcotθ
Can you simplify further?

I see that you're getting a common denominator on the left side, but I don't see where I can go from there.

Is it actually possible to simplify it further, or are you saying there is some identity that I'm not seeing?

Thanks!

Originally Posted by JennyFlowers

I see that you're getting a common denominator on the left side, but I don't see where I can go from there.

Is it actually possible to simplify it further, or are you saying there is some identity that I'm not seeing?

Thanks!

Can you write down all three identities?
And what is the relation between tanθ and cotθ?

Do you mean reciprocal, quotient and Pythagorean identities?

The relationship of tan θ and cot θ is that they are reciprocals of one another.

So... tan θ is equivalent to 1 / cot θ and vice versa.

But I'm still not seeing how to apply that here!

Originally Posted by JennyFlowers

Do you mean reciprocal, quotient and Pythagorean identities?

The relationship of tan θ and cot θ is that they are reciprocals of one another.

So... tan θ is equivalent to 1 / cot θ and vice versa.

But I'm still not seeing how to apply that here!

What is tanθ*cotθ?
What is sin^2(θ) + cos^2(θ) = ?
Similarly can you write down other two relations? If you can't, please open any trigonometry book and search for that.

tan θ * cot θ = 1 (Right?)

sin^2(θ) + cos^2(θ) = 1

I suppose the other two relations you're talking about are:

1 + tan^2(θ) = sec^2(θ)

and

1 + cot^2(θ) = csc^2(θ)

This seems interesting, but I'm not putting this together for the complete picture.

Okay, so now I have:

tan^2(θ) - cot^2(θ) = sec^2(θ) - cos^2(θ)

I'm not sure where to go from here.

tan^2(θ) - cot^2(θ) = sec^2(θ) - cos^2(θ)


Check the last term.

Check it?

I'm not sure what you mean.

Do you mean that it's 1 / sec^2(θ)?

I understand that we are subtracting reciprocals on both sides, but I'm not aware of any properties that help when subtracting reciprocals.

Kind of lost, here...

1 + tan^2(θ) = sec^2(θ)
and
1 + cot^2(θ) = csc^2(θ)

You have written these two relations.
From these equations
what is tan^2θ and cot^2θ?
Then what is
tan^2θ - cot^2θ?
What is required to show in the given problem?

Originally Posted by sa-ri-ga-ma

1 + tan^2(θ) = sec^2(θ)
and
1 + cot^2(θ) = csc^2(θ)

You have written these two relations.
From these equations
what is tan^2θ and cot^2θ?
Then what is
tan^2θ - cot^2θ?
What is required to show in the given problem?

tan^2θ = sec^2θ - 1
cot^2θ = csc^2θ - 1

tan^2θ - cot^2θ = sec^2θ - 1 - (csc^2θ - 1) = sec^2θ - 1 - csc^2θ + 1 = sec^2θ - csc^2θ

Is that right?

Originally Posted by JennyFlowers

tan^2θ = sec^2θ - 1
cot^2θ = csc^2θ - 1

tan^2θ - cot^2θ = sec^2θ - 1 - (csc^2θ - 1) = sec^2θ - 1 - csc^2θ + 1 = sec^2θ - csc^2θ

Is that right?

Yes.