# Prove that tan20+tan80+tan140=3sqrt3?

Jul 23, 2018

#### Explanation:

We take ,

$L H S = \tan {20}^{\circ} + \tan {80}^{\circ} + \tan {140}^{\circ}$

$\textcolor{w h i t e}{L H S} = \tan {20}^{\circ} + \tan \left({60}^{\circ} + {20}^{\circ}\right) + \tan \left({120}^{\circ} + {20}^{\circ}\right)$

$\textcolor{w h i t e}{L H S}$=$\tan {20}^{\circ} + \frac{\tan {60}^{\circ} + \tan {20}^{\circ}}{1 - \tan {60}^{\circ} \tan {20}^{\circ}} + \frac{\tan {120}^{\circ} + \tan {20}^{\circ}}{1 - \tan {120}^{\circ} \tan {20}^{\circ}}$

Subst. color(blue)(tan60^circ=sqrt3 ,tan120^circ=-sqrt3 and tan20^circ=t

$L H S = t + \frac{\sqrt{3} + t}{1 - \sqrt{3} t} + \frac{- \sqrt{3} + t}{1 + \sqrt{3} t}$

$\textcolor{w h i t e}{L H S} = t + \frac{\left(\sqrt{3} + t\right) \left(1 + \sqrt{3} t\right) + \left(- \sqrt{3} + t\right) \left(1 - \sqrt{3} t\right)}{\left(1 - \sqrt{3} t\right) \left(1 + \sqrt{3} t\right)}$

$\textcolor{w h i t e}{L H S} = t + \frac{\sqrt{3} + 3 t + t + \sqrt{3} {t}^{2} - \sqrt{3} + 3 t + t - \sqrt{3} {t}^{2}}{1 - 3 {t}^{2}}$

$\textcolor{w h i t e}{L H S} = t + \frac{8 t}{1 - 3 {t}^{2}}$

$\textcolor{w h i t e}{L H S} = \frac{t - 3 {t}^{3} + 8 t}{1 - 3 {t}^{2}}$

$\textcolor{w h i t e}{L H S} = \frac{9 t - 3 {t}^{3}}{1 - 3 {t}^{2}}$

color(white)(LHS)=3[(3t-t^3)/(1-3t^2)]towhere,color(blue)(t=tan20^circ

$\textcolor{w h i t e}{L H S} = 3 \left[\frac{3 \tan {20}^{\circ} - {\tan}^{3} {20}^{\circ}}{1 - 3 {\tan}^{2} {20}^{\circ}}\right]$

$\textcolor{w h i t e}{L H S} = 3 \left[\tan 3 \left({20}^{\circ}\right)\right] \to A p p l y \left(2\right)$ for $\theta = {20}^{\circ}$

$L H S = 3 \tan {60}^{\circ}$

$L H S = 3 \sqrt{3} = R H S$

Note :

$\left(1\right) \tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\left(2\right) \tan 3 \theta = \frac{3 \tan \theta - {\tan}^{3} \theta}{1 - 3 {\tan}^{2} \theta}$

Jul 25, 2018

$L H S = \tan 20 + \tan 80 + \tan 140$

$= \tan 20 + \tan 80 + \tan \left(180 - 40\right)$

$= \tan 20 + \tan 80 - \tan 40$

$= \tan 20 + \sin \frac{80}{\cos} 80 - \sin \frac{40}{\cos} 40$

$= \sin \frac{20}{\cos} 20 + \frac{\sin 80 \cos 40 - \cos 80 \sin 40}{\cos 80 \cos 40}$

$= \frac{\sin 20 \cos 80 \cos 40 + \sin 40 \cos 20}{\cos 20 \cos 80 \cos 40}$

Now denominator of this expression

$= \cos 20 \cos 80 \cos 40$

$= \frac{4 \cdot 2 \sin 20 \cos 20 \cos 40 \cos 80}{8 \sin 20}$

$= \frac{2 \cdot 2 \sin 40 \cos 40 \cos 80}{8 \sin 20}$

$= \frac{2 \sin 80 \cos 80}{8 \sin 20}$

$= \frac{\sin 160}{8 \sin 20}$

$= \frac{\sin \left(180 - 20\right)}{8 \sin 20}$

$= \frac{\sin 20}{8 \sin 20}$

$= \frac{1}{8}$

Hence

$L H S = 8 \left(\sin 20 \cos 80 \cos 40 + \sin 40 \cos 20\right)$

$= 4 \sin 20 \cdot \left(2 \cos 80 \cos 40\right) + 4 \cdot 2 \sin 40 \cos 20$

$= 4 \sin 20 \left(\cos 120 + \cos 40\right) + 4 \left(\sin 60 + \sin 20\right)$

$= 4 \sin 20 \left(- \frac{1}{2} + \cos 40\right) + 4 \left(\frac{\sqrt{3}}{2} + \sin 20\right)$

$= - 2 \sin 20 + 4 \sin 20 \cos 40 + 2 \sqrt{3} + 4 \sin 20$

$= 4 \sin 20 \cos 40 + 2 \sqrt{3} + 2 \sin 20$

$= 2 \left(\sin 60 - \sin 20\right) + 2 \sqrt{3} + 2 \sin 20$

$= 2 \left(\frac{\sqrt{3}}{2} - \sin 20\right) + 2 \sqrt{3} + 2 \sin 20$

$= \sqrt{3} - 2 \sin 20 + 2 \sqrt{3} + 2 \sin 20$

$= 3 \sqrt{3}$

Jul 25, 2018

A funny approach utilising the anwer $3 \sqrt{3}$ given.

We can write LHS as follows as we know $\sqrt{3} = \tan 60$

$L H S = \tan 20 + \tan 80 + \tan 140$

$= 3 \sqrt{3} + \left(\tan 20 - \tan 60\right) + \left(\tan 80 - \tan 60\right) + \left(\tan 140 - \tan 60\right)$

$= 3 \sqrt{3} + \left(\tan 20 - \tan 60\right) + \left(\tan 80 - \tan 60\right) + \left(\tan \left(180 - 40\right) - \tan 60\right)$

$= 3 \sqrt{3} + \left(\tan 20 - \tan 60\right) + \left(\tan 80 - \tan 60\right) - \left(\tan 40 + \tan 60\right)$

$= 3 \sqrt{3} + \left(\sin \frac{20}{\cos} 20 - \sin \frac{60}{\cos} 60\right) + \left(\sin \frac{80}{\cos} 80 - \sin \frac{60}{\cos} 60\right) - \left(\sin \frac{40}{\cos} 40 + \sin \frac{60}{\cos} 60\right)$

$= 3 \sqrt{3} - \sin \frac{60 - 20}{\cos 20 \cos 60} + \sin \frac{80 - 60}{\cos 80 \cos 60} - \sin \frac{60 + 40}{\cos 40 \cos 60}$

$= 3 \sqrt{3} - \frac{2 \sin 40}{\cos} 20 + \frac{2 \sin 20}{\cos} 80 - \frac{2 \sin 100}{\cos} 40$

$= 3 \sqrt{3} - \frac{4 \sin 20 \cos 20}{\cos} 20 + \frac{4 \sin 10 \cos 10}{\sin} 10 - \frac{4 \sin 40 \cos 40}{\cos} 40$

$= 3 \sqrt{3} - 4 \sin 20 + 4 \cos 10 - 4 \sin 40$

$= 3 \sqrt{3} - 4 \left(\sin 20 + \sin 40\right) + 4 \cos 10$

$= 3 \sqrt{3} - 4 \left(2 \sin 30 \cos 1 0\right) + 4 \cos 10$

$= 3 \sqrt{3} - 4 \left(2 \cdot \frac{1}{2} \cdot \cos 1 0\right) + 4 \cos 10$

$= 3 \sqrt{3} - 4 \cos 10 + 4 \cos 10$

$= 3 \sqrt{3}$

Jul 26, 2018

Explanation in below

#### Explanation:

$x = \tan 20 + \tan 80 + \tan 140$

=$\sin \frac{20}{\cos} 20 + \sin \frac{80}{\cos} 80 + \tan \left(180 - 40\right)$

=$\frac{\cos 80 \cdot \sin 20 + \sin 80 \cdot \cos 20}{\cos 80 \cdot \cos 20} - \tan 40$

=$\sin \frac{80 + 20}{\cos 80 \cdot \cos 20} - \sin \frac{40}{\cos} 40$

=$\sin \frac{100}{\cos 80 \cdot \cos 20} - \sin \frac{40}{\cos} 40$

=$\sin \frac{80}{\cos 80 \cdot \cos 20} - \sin \frac{40}{\cos} 40$

=$\frac{\sin 80 \cdot \cos 40 - \cos 80 \cdot \sin 40 \cdot \cos 20}{\cos 80 \cdot \cos 40 \cdot \cos 20}$

=$\frac{\sin 20 \cdot \left(8 \sin 80 \cdot \cos 40 - 8 \cos 80 \cdot \sin 40 \cdot \cos 20\right)}{8 \cos 80 \cdot \cos 40 \cdot \cos 20 \cdot \sin 20}$

=$\frac{\sin 20 \cdot \left(4 \sin 120 + 4 \sin 40 - 4 \cos 20 \cdot \left(\sin 120 - \sin 40\right)\right)}{4 \cos 80 \cdot \cos 40 \cdot \sin 40}$

=$\frac{\sin 20 \cdot \left(4 \sin 120 + 4 \sin 40 - 4 \sin 120 \cdot \cos 20 + 4 \sin 40 \cdot \cos 20\right)}{2 \cos 80 \cdot \sin 80}$

=$\frac{\sin 20 \cdot \left(4 \sin 60 + 4 \sin 40 - 4 \sin 60 \cdot \cos 20 + 4 \sin 40 \cdot \cos 20\right)}{\sin 160}$

=$\frac{\sin 20 \cdot \left(4 \sin 60 + 4 \sin 40 - 2 \sin 80 - 2 \sin 40 + 2 \sin 60 + 2 \sin 20\right)}{\sin 20}$

=$6 \sin 60 + 2 \sin 40 - 2 \sin 80 + 2 \sin 20$

=$3 \sqrt{3} + 2 \sin 20 - \left(2 \sin 80 - 2 \sin 40\right)$

=$3 \sqrt{3} + 2 \sin 20 - 4 \cos 60 \cdot \sin 20$

=$3 \sqrt{3} + 2 \sin 20 - 2 \sin 20$

=$3 \sqrt{3}$