# Math Help - [SOLVED] exact answer using trigonometric difference identity

1. ## [SOLVED] exact answer using trigonometric difference identity

I'm afraid I'm lost attempting to solve what is probably a really simple problem. The idea is to get an exact answer using the trigonometric difference identity for tangent.

$\tan(\frac{-\pi}{12})$

which is the same as
$\tan(\frac{\pi}{3} - \frac{\pi}{4})$

the formula is

$\frac{tanA - tanB}{1+tanAtanB}$
thus
$\frac{\tan(\frac{pi}{3}) - \tan(\frac{pi}{4})}{1+\tan(\frac{pi}{3})tan(\frac{ pi}{4})}$
ergo
$\frac{\sqrt3 - 1}{1+\sqrt3}$

and this is where I'm stuck. The book has the answer as $-2 + \sqrt3$, so either I'm completely wrong, or I just don't know how to simplify properly. Any assistance would be greatly appreciated.

2. Originally Posted by satis
I'm afraid I'm lost attempting to solve what is probably a really simple problem. The idea is to get an exact answer using the trigonometric difference identity for tangent.

$\tan(\frac{-\pi}{12})$

which is the same as
$\color {red} \tan(\frac{\pi}{3} - \frac{\pi}{4}) \quad$\\edit: $\color {blue} \tan(\frac{\pi}{4}-\frac{\pi}{3})$
the formula is

$\frac{tanA - tanB}{1+tanAtanB}$
thus
$\frac{\tan(\frac{\pi}{3}) - \tan(\frac{\pi}{4})}{1+\tan(\frac{\pi}{3})tan(\fra c{\pi}{4})}$
ergo
$\frac{\sqrt3 - 1}{1+\sqrt3} \quad$ \\edit $\color {blue}\frac{1- \sqrt3}{1+\sqrt3} \quad$

and this is where I'm stuck. The book has the answer as $-2 + \sqrt3$, so either I'm completely wrong, or I just don't know how to simplify properly. Any assistance would be greatly appreciated.
rationalize $\color {blue}\frac{1- \sqrt3}{1+\sqrt3} \quad$
hint : $\frac{1- \sqrt3}{1+\sqrt3} \times \frac{1- \sqrt3}{1-\sqrt3}$

3. Originally Posted by satis
I'm afraid I'm lost attempting to solve what is probably a really simple problem. The idea is to get an exact answer using the trigonometric difference identity for tangent.

$\tan(\frac{-\pi}{12})$

which is the same as
$\tan(\frac{\pi}{3} - \frac{\pi}{4})$

the formula is

$\frac{tanA - tanB}{1+tanAtanB}$
thus
$\frac{\tan(\frac{pi}{3}) - \tan(\frac{pi}{4})}{1+\tan(\frac{pi}{3})tan(\frac{ pi}{4})}$
ergo
$\frac{\sqrt3 - 1}{1+\sqrt3}$

and this is where I'm stuck. The book has the answer as $-2 + \sqrt3$, so either I'm completely wrong, or I just don't know how to simplify properly. Any assistance would be greatly appreciated.
Multiply the numerator and denominator of your answer by the conjugate surd $1 - \sqrt{3}$ and simplify to what the book has.

4. Thank you both for your assistance. I realize (now that it's been pointed out) I made a mistake right from the beginning. And rationalizing still poses me some difficulties, apparently, even though that's really more algebra than trig. However, I was able to work it out thanks to the feedback. Once again, thanks... I'm feeling much less lost now.