# Verifying that this is an identity?

• Aug 5th 2011, 12:58 PM
explodingtoenails
Verifying that this is an identity?
Trig and I are not very good friends...

$\displaystyle \displaystyle \frac{1 + tan^2x}{csc^2x} = tan^2x$

This is what I tried:

$\displaystyle \displaystyle \frac{1 + (sin^2x/cos^2x)}{(1/sin^2x)} = tan^2x$

$\displaystyle \displaystyle \frac{cos^2x + sin^2x}{cos^2x} * \frac{sin^2x}{1} = tan^2x$

$\displaystyle (sin^2x)(sin^2x) = tan^2x$

I don't know why it's not equaling up. D:
• Aug 5th 2011, 01:03 PM
TheChaz
Re: Verifying that this is an identity?
Quote:

Originally Posted by explodingtoenails
Trig and I are not very good friends...

$\displaystyle \displaystyle \frac{1 + tan^2x}{csc^2x} = tan^2x$

This is what I tried:

$\displaystyle \displaystyle \frac{1 + (sin^2x/cos^2x)}{(1/sin^2x)} = tan^2x$

$\displaystyle \displaystyle \frac{cos^2x + sin^2x}{cos^2x} * \frac{sin^2x}{1} = tan^2x$

...

I don't know why it's not equaling up. D:

From here, the top left is ONE.
Then you have (sin/cos)^2
• Aug 5th 2011, 01:06 PM
explodingtoenails
Re: Verifying that this is an identity?
Thank you so much dude. I didn't know that cos^2x + sin^2x = 1.
• Aug 5th 2011, 01:42 PM
TheChaz
Re: Verifying that this is an identity?
Quote:

Originally Posted by explodingtoenails
Thank you so much dude. I didn't know that cos^2(x) + sin^2(x) = 1.

That's kind of important! From it, many (most?) identities are derived.
• Aug 15th 2011, 07:55 PM
Drexel28
Re: Verifying that this is an identity?
Quote:

Originally Posted by TheChaz
That's kind of important! From it, many (most?) identities are derived.

The most important identity really. By this I mean that as rings $\displaystyle \mathbb{R}[x,y]/(x^2-y^2-1)\cong \mathbb{R}[\cos(\theta),\sin(\theta)]$. In other words, if you take the ring $\displaystyle \mathbb{R}[x,y]$ and impose the conditions that $\displaystyle x^2+y^2=1$ then you really just have the ring of trigonometric polynomials $\displaystyle \mathbb{R}[\cos(\theta),\sin(\theta)]$.