1. ## trig identity help.

Write the following in their simplest form, involving only one trigonometric function:

$\frac{tan \theta}{sec^{2} \theta-2}$
$1 + tan^{2}\theta = sec^{2} \theta$

$\frac{tan \theta}{ 1 + tan^{2} \theta - 2}$

$\frac{tan\theta}{tan^2 \theta-1}$

I am stuck at this point, I know I have to get it to look like the $tan2\theta$ expansion,

any hints/help appreciated.

2. Originally Posted by Tweety
$1 + tan^{2}\theta = sec^{2} \theta$

$\frac{tan \theta}{ 1 + tan^{2} \theta - 2}$

$\frac{tan\theta}{tan^2 \theta-1}$

I am stuck at this point, I know I have to get it to look like the $tan2\theta$ expansion,

any hints/help appreciated.
Hi Tweety,

Maybe you could do something like this.

$\frac{tan\theta}{tan^2 \theta-1} = -\frac{\tan \theta}{1-\tan^2 \theta} \cdot \frac{2}{2}=-\frac{2\tan \theta}{2(1-\tan^2 \theta)}=-\frac{1}{2} \tan 2 \theta$

3. Originally Posted by masters
Hi Tweety,

Maybe you could do something like this.

$\frac{tan\theta}{tan^2 \theta-1} = -\frac{\tan \theta}{1-\tan^2 \theta} \cdot \frac{2}{2}=-\frac{2\tan \theta}{2(1-\tan^2 \theta)}=-\frac{1}{2} \tan 2 \theta$
thanks,

Just one question. How did you get $1-tan^{2} \theta$ from $tan^{2} \theta - 1$?

Did you multiply top and bottom by minus one?

4. Originally Posted by Tweety

Did you multiply top and bottom by minus one?