اگر $\tan \left( ۲a+\frac{B}{۳} \right)=\sqrt{۳}+۱$ و $\tan \left( ۲a-\frac{B}{۳} \right)=\sqrt{۳}-۱$ باشد، $\tan \frac{۲B}{۳}$ برابر است با:
$\frac{2B}{3}=\left( 2a+\frac{B}{3} \right)-\left( 2a-\frac{B}{3} \right)$ $\tan \frac{2B}{3}=\frac{\tan \left( 2a+\frac{B}{3} \right)-\tan \left( 2a-\frac{B}{3} \right)}{1+\tan \left( 2a+\frac{B}{3} \right)\tan \left( 2a-\frac{B}{3} \right)}=\frac{\sqrt{3}+1-\sqrt{3}+1}{1+\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}=\frac{2}{1+\left( 3-1 \right)}=\frac{2}{3}$