# Could you demonstrate this equation?! And thanks you btw! 3arctan(2-sqr(3))=arctan1/2 + arctan1/3

Mar 12, 2018

Use, $\left(2 - \sqrt{3}\right) = \tan {15}^{0}$ to prove the equation.

#### Explanation:

Here, $3 {\tan}^{- 1} \left(2 - \sqrt{3}\right) = {\tan}^{-} 1 \left(\frac{1}{2}\right) + {\tan}^{-} 1 \left(\frac{1}{3}\right)$
We know that,$\textcolor{red}{\tan {15}^{0}} = \tan \left({60}^{0} - {45}^{0}\right)$
=>tan15^0=(tan60^0-tan45^0)/(1+tan60^0tan45^0
=>tan15^0=(sqrt(3)-1)/(1+sqrt3)=((sqrt(3)-1)(sqrt(3)-1))/((sqrt(3)+1)(sqrt(3)-1))=(3-2sqrt(3)+1)/(3-1)=(4-2sqrt3)/2=color(red)((2-sqrt3)
$L H S = 3 {\tan}^{- 1} \left(2 - \sqrt{3}\right) = 3 {\tan}^{1} \left(\tan {15}^{0}\right) = 3 \cdot {15}^{0} = {45}^{0}$
$\textcolor{b l u e}{\implies L H S = \frac{\pi}{4}}$
$R H S = {\tan}^{-} 1 \left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}\right)$
$= {\tan}^{-} 1 \left(\frac{\frac{3 + 2}{6}}{\frac{6 - 1}{6}}\right)$
$= {\tan}^{-} 1 \left(\frac{5}{5}\right) = {\tan}^{-} 1 \left(1\right)$
$\textcolor{b l u e}{R H S = \frac{\pi}{4}}$