# Thread: Sum and Difference Formulas for Tangents

1. ## Sum and Difference Formulas for Tangents

I've been having a lot of trouble understanding how to use the sum and difference formulas for tangents. I can never quite simplify the question completely.

Sample:

Find the exact value of tan(19pi/12)

2. Originally Posted by magentarita
I've been having a lot of trouble understanding how to use the sum and difference formulas for tangents. I can never quite simplify the question completely.

Sample:

Find the exact value of tan(19pi/12)
$\tan \left(\alpha+\beta\right) = \frac{\tan \alpha+\tan \beta}{1-\tan\alpha\tan\beta}$

$\frac{19\pi}{12} = \frac{4\pi}{3} + \frac{\pi}{4}$

$\therefore \tan \left( \frac{19 \pi}{12} \right) = \tan \left(\frac{4\pi}{3} + \frac{\pi}{4}\right)$

$=\frac{\tan
\left(\frac{4\pi}{3}\right)
+\tan \left(\frac{\pi}{4}\right)}{1-\tan
\left(\frac{4\pi}{3}\right)
\tan \left(\frac{\pi}{4}\right)} = \frac{\sqrt{3}+1}{1-\sqrt{3}}$

To simplify $\frac{\sqrt{3}+1}{1-\sqrt{3}}$, multiply numerator and the denominator by the conjugate of the denominator which is $1+\sqrt{3}$.

Originally Posted by Air
$\tan \left(\alpha+\beta\right) = \frac{\tan \alpha+\tan \beta}{1-\tan\alpha\tan\beta}$

$\frac{19\pi}{12} = \frac{4\pi}{3} + \frac{\pi}{4}$

$\therefore \tan \left( \frac{19 \pi}{12} \right) = \tan \left(\frac{4\pi}{3} + \frac{\pi}{4}\right)$

$=\frac{\tan
\left(\frac{4\pi}{3}\right)
+\tan \left(\frac{\pi}{4}\right)}{1-\tan
\left(\frac{4\pi}{3}\right)
\tan \left(\frac{\pi}{4}\right)} = \frac{\sqrt{3}+1}{1-\sqrt{3}}$

To simplify $\frac{\sqrt{3}+1}{1-\sqrt{3}}$, multiply numerator and the denominator by the conjugate of the denominator which is $1+\sqrt{3}$.
Air, your replies are simply amazing!!