# Thread: verifying identities attempt 3...

1. ## verifying identities attempt 3...

Here is the problem:

$\frac{tan-cot}{tan+cot}+1 = 2sin^{2}$

I convert it into terms of cos and sin and try to simplify the left side.

$
\frac{sin}{cos}-\frac{cos}{sin}$

_________

$\frac{sin}{cos}+\frac{cos}{sin}$

Then, if I took it a step further and made them get a denominator of cos sin they'd end up on top with

$sin^{2}-cos^{2}$ and on the bottom $sin^{2}+cos^{2}$. I could make the bottom (using the Pythagorean identity):

$\frac{1}{cos sin}$

And that's how far I've gotten before everything fails.

2. Originally Posted by A Beautiful Mind
Here is the problem:

$\frac{tan-cot}{tan+cot}+1 = 2sin^{2}$

I convert it into terms of cos and sin and try to simplify the left side.

$
\frac{sin}{cos}-\frac{cos}{sin}$

_________

$\frac{sin}{cos}+\frac{cos}{sin}$

Then, if I took it a step further and made them get a denominator of cos sin they'd end up on top with

$sin^{2}-cos^{2}$ and on the bottom $sin^{2}+cos^{2}$. I could make the bottom (using the Pythagorean identity):

$\frac{1}{cos sin}$

And that's how far I've gotten before everything fails.

So far it sounds like you've ended up with

$\frac{\sin^2{x} - \cos^2{x}}{\sin^2{x} + \cos^2{x}} + 1$

$= \frac{\sin^2{x} - (1 - \sin^2{x})}{1} + 1$

$= \sin^2{x} - 1 + \sin^2{x} + 1$

$= 2\sin^2{x}$.