proving tangent formula
(tangent squared theta = 1-cos squared theta over 1+cos sqared theta)

Do you mean?.

\text{tan}^2\theta \ =\ \frac{1\ -\ \text{\large{cos}}^2\theta}{1\ +\ \text{\large{cos}}^2\theta}

If so, try evaluating both sides of the proposed equality for a random value of theta.

Are you sure this formula is correct? I can't seem to get it. I tried several ways, no success.

Agreed... the way I structured the equation it's definitely NOT an identity. That's why I asked Paola to run a simple 'test' with any random pick for theta... just about any angle measure would provide the counterexample to the assertion that it's an ID.

On the other hand, maybe Paola intended a different 'setup' of the equation. Just have to stand by for that.

Cheers

I also tried tan^2\theta = 1 - \frac{cos^2\theta}{1+cos^2\theta}, but to no avail...

I also tried tan^2\theta = 1 - \frac{cos^2\theta}{1+cos^2\theta}, but to no avail...

Do you think it could be:


tan^2\theta = \frac{1-cos^2\theta}{cos^2\theta}

That should make it easily solvable.

Yes, that is true... I wonder where the OP got that "1+" up there...