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

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

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Answer 1 · 2018-03-12T10:15:20

Use, $(2 - \sqrt{3}) = {tan 15}^{0}$ $(2-sqrt3)=tan15^0$ to prove the equation.

Explanation:

Here, $3 {tan}^{- 1} (2 - \sqrt{3}) = {tan}^{- 1} (\frac{1}{2}) + {tan}^{- 1} (\frac{1}{3})$ $3tan^(-1)(2-sqrt3)=tan^-1(1/2)+tan^-1(1/3)$
We know that, ${tan 15}^{0} = tan (60^{0} - 45^{0})$ $color(red)(tan15^0)=tan(60^0-45^0)$
$\Rightarrow {tan 15}^{0} = \frac{{tan 60}^{0} - {tan 45}^{0}}{1 + {tan 60}^{0} {tan 45}^{0}}$ $=>tan15^0=(tan60^0-tan45^0)/(1+tan60^0tan45^0$
$\Rightarrow {tan 15}^{0} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1) (\sqrt{3} - 1)}{(\sqrt{3} + 1) (\sqrt{3} - 1)} = \frac{3 - 2 \sqrt{3} + 1}{3 - 1} = \frac{4 - 2 \sqrt{3}}{2} = (2 - \sqrt{3})$ $=>tan15^0=(sqrt(3)-1)/(1+sqrt3)=((sqrt(3)-1)(sqrt(3)-1))/((sqrt(3)+1)(sqrt(3)-1))=(3-2sqrt(3)+1)/(3-1)=(4-2sqrt3)/2=color(red)((2-sqrt3)$
$L H S = 3 {tan}^{- 1} (2 - \sqrt{3}) = 3 {tan}^{1} ({tan 15}^{0}) = 3 \cdot 15^{0} = 45^{0}$ $LHS=3tan^(-1)(2-sqrt3)=3tan^1(tan15^0)=3*15^0=45^0$
$\Rightarrow L H S = \frac{π}{4}$ $color(blue)(=>LHS=pi/4)$
$R H S = {tan}^{- 1} (\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}})$ $RHS=tan^-1((1/2+1/3)/(1-1/2*1/3))$
$= {tan}^{- 1} (\frac{\frac{3 + 2}{6}}{\frac{6 - 1}{6}})$ $=tan^-1(((3+2)/6)/((6-1)/6))$
$= {tan}^{- 1} (\frac{5}{5}) = {tan}^{- 1} (1)$ $=tan^-1(5/5)=tan^-1(1)$
$R H S = \frac{π}{4}$ $color(blue)(RHS=pi/4)$

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

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