Prove that $tan 20 + tan 80 + tan 140 = 3 \sqrt{3}$ $tan20+tan80+tan140=3sqrt3$ ?

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Prove that $tan 20 + tan 80 + tan 140 = 3 \sqrt{3}$ $tan20+tan80+tan140=3sqrt3$ ?

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Answer 1 · 2018-07-23T17:46:00

Please see below.

Explanation:

We take ,

$L H S = {tan 20}^{\circ} + {tan 80}^{\circ} + {tan 140}^{\circ}$ $LHS=tan 20^circ+tan80^circ+tan140^circ$

$L H S = {tan 20}^{\circ} + tan (60^{\circ} + 20^{\circ}) + tan (120^{\circ} + 20^{\circ})$ $color(white)(LHS)=tan20^circ+tan(60^circ+20^circ)+tan(120^circ+20^circ)$

$L H S$ $color(white)(LHS)$ = ${tan 20}^{\circ} + \frac{{tan 60}^{\circ} + {tan 20}^{\circ}}{1 - {tan 60}^{\circ} {tan 20}^{\circ}} + \frac{{tan 120}^{\circ} + {tan 20}^{\circ}}{1 - {tan 120}^{\circ} {tan 20}^{\circ}}$ $tan20^circ+(tan60^circ+tan20^circ)/(1-tan60^circtan20^circ)+(tan120^circ+tan20^circ)/(1-tan120^circtan20^circ)$

Subst. ${tan 60}^{\circ} = \sqrt{3}, {tan 120}^{\circ} = - \sqrt{3} and {tan 20}^{\circ} = t$ $color(blue)(tan60^circ=sqrt3 ,tan120^circ=-sqrt3 and tan20^circ=t$

$L H S = t + \frac{\sqrt{3} + t}{1 - \sqrt{3} t} + \frac{- \sqrt{3} + t}{1 + \sqrt{3} t}$ $LHS=t+(sqrt3+t)/(1-sqrt3t)+(-sqrt3+t)/(1+sqrt3t)$

$L H S = t + \frac{(\sqrt{3} + t) (1 + \sqrt{3} t) + (- \sqrt{3} + t) (1 - \sqrt{3} t)}{(1 - \sqrt{3} t) (1 + \sqrt{3} t)}$ $color(white)(LHS)=t+{(sqrt3+t)(1+sqrt3t)+(-sqrt3+t)(1-sqrt3t))/((1-sqrt3t)(1+sqrt3t))$

$L H S = t + \frac{\sqrt{3} + 3 t + t + \sqrt{3} t^{2} - \sqrt{3} + 3 t + t - \sqrt{3} t^{2}}{1 - 3 t^{2}}$ $color(white)(LHS)=t+(sqrt3+3t+t+sqrt3t^2-sqrt3+3t+t-sqrt3t^2)/(1-3t^2)$

$L H S = t + \frac{8 t}{1 - 3 t^{2}}$ $color(white)(LHS)=t+(8t)/(1-3t^2)$

$L H S = \frac{t - 3 t^{3} + 8 t}{1 - 3 t^{2}}$ $color(white)(LHS)=(t-3t^3+8t)/(1-3t^2)$

$L H S = \frac{9 t - 3 t^{3}}{1 - 3 t^{2}}$ $color(white)(LHS)=(9t-3t^3)/(1-3t^2)$

$L H S = 3 [\frac{3 t - t^{3}}{1 - 3 t^{2}}] \to w h e r e, t = {tan 20}^{\circ}$ $color(white)(LHS)=3[(3t-t^3)/(1-3t^2)]towhere,color(blue)(t=tan20^circ$

$L H S = 3 [\frac{3 {tan 20}^{\circ} - {tan}^{3} 20^{\circ}}{1 - 3 {tan}^{2} 20^{\circ}}]$ $color(white)(LHS)=3[(3tan20^circ-tan^3 20^circ)/(1-3tan^2 20^circ)]$

$L H S = 3 [tan 3 (20^{\circ})] \to A p p l y (2)$ $color(white)(LHS)=3[tan3(20^circ)]toApply(2)$ for $θ = 20^{\circ}$ $theta=20^circ$

$L H S = 3 {tan 60}^{\circ}$ $LHS=3tan60^circ$

$L H S = 3 \sqrt{3} = R H S$ $LHS=3sqrt3=RHS$

Note :

$(1) tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ $(1) tan(A+B)=(tanA+tanB)/(1-tanAtanB)$

$(2) tan 3 θ = \frac{3 tan θ - {tan}^{3} θ}{1 - 3 {tan}^{2} θ}$ $(2)tan3theta=(3tantheta-tan^3theta)/(1-3tan^2theta)$

Answer 2 · 2018-07-25T04:39:01

$L H S = tan 20 + tan 80 + tan 140$ $LHS=tan20+tan80+tan140$

$= tan 20 + tan 80 + tan (180 - 40)$ $=tan20+tan80+tan(180-40)$

$= tan 20 + tan 80 - tan 40$ $=tan20+tan80-tan 40$

$= tan 20 + \frac{sin 80}{cos 80} - \frac{sin 40}{cos 40}$ $=tan20+sin 80/cos 80-sin 40/cos 40$

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

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

Now denominator of this expression

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

$= \frac{4 \cdot 2 sin 20 cos 20 cos 40 cos 80}{8 sin 20}$ $=(4*2sin 20cos 20cos 40cos 80)/(8sin 20)$

$= \frac{2 \cdot 2 sin 40 cos 40 cos 80}{8 sin 20}$ $=(2*2sin 40cos 40cos 80)/(8sin 20)$

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

$= \frac{sin 160}{8 sin 20}$ $=(sin 160)/(8sin 20)$

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

$= \frac{sin 20}{8 sin 20}$ $=(sin 20)/(8sin 20)$

$= \frac{1}{8}$ $=1/8$

Hence

$L H S = 8 (sin 20 cos 80 cos 40 + sin 40 cos 20)$ $LHS=8(sin 20cos 80cos 40+sin 40cos 20)$

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

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

$= 4 sin 20 (- \frac{1}{2} + cos 40) + 4 (\frac{\sqrt{3}}{2} + sin 20)$ $=4sin 20(-1/2+cos 40)+4(sqrt3/2+sin 20)$

$= - 2 sin 20 + 4 sin 20 cos 40 + 2 \sqrt{3} + 4 sin 20$ $=-2sin 20+4sin 20cos 40+2sqrt3+4sin 20$

$= 4 sin 20 cos 40 + 2 \sqrt{3} + 2 sin 20$ $=4sin 20cos 40+2sqrt3+2sin 20$

$= 2 (sin 60 - sin 20) + 2 \sqrt{3} + 2 sin 20$ $=2(sin 60-sin 20)+2sqrt3+2sin 20$

$= 2 (\frac{\sqrt{3}}{2} - sin 20) + 2 \sqrt{3} + 2 sin 20$ $=2(sqrt3/2-sin 20)+2sqrt3+2sin 20$

$= \sqrt{3} - 2 sin 20 + 2 \sqrt{3} + 2 sin 20$ $= sqrt3-2sin 20+2sqrt3+2sin 20$

$= 3 \sqrt{3}$ $=3sqrt3$

Answer 3 · 2018-07-25T12:17:51

A funny approach utilising the anwer $3 \sqrt{3}$ $3sqrt3$ given.

We can write LHS as follows as we know $\sqrt{3} = tan 60$ $sqrt3=tan 60$

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

$= 3 \sqrt{3} + (tan 20 - tan 60) + (tan 80 - tan 60) + (tan 140 - tan 60)$ $=3sqrt3+(tan 20-tan 60)+(tan 80-tan 60) +(tan 140-tan 60)$

$= 3 \sqrt{3} + (tan 20 - tan 60) + (tan 80 - tan 60) + (tan (180 - 40) - tan 60)$ $=3sqrt3+(tan 20-tan 60)+(tan 80-tan 60) +(tan (180-40)-tan 60)$

$= 3 \sqrt{3} + (tan 20 - tan 60) + (tan 80 - tan 60) - (tan 40 + tan 60)$ $=3sqrt3+(tan 20-tan 60)+(tan 80-tan 60) -(tan 40+tan 60)$

$= 3 \sqrt{3} + (\frac{sin 20}{cos 20} - \frac{sin 60}{cos 60}) + (\frac{sin 80}{cos 80} - \frac{sin 60}{cos 60}) - (\frac{sin 40}{cos 40} + \frac{sin 60}{cos 60})$ $=3sqrt3+(sin 20/cos 20-sin 60/cos60)+(sin 80/cos 80-sin 60/cos60 )-(sin 40/cos40+sin 60/cos60)$

$= 3 \sqrt{3} - \frac{sin (60 - 20)}{cos 20 cos 60} + \frac{sin (80 - 60)}{cos 80 cos 60} - \frac{sin (60 + 40)}{cos 40 cos 60}$ $=3sqrt3-sin(60- 20)/(cos 20cos60)+sin (80-60)/(cos 80cos60 )-sin (60+40)/(cos40cos60)$

$= 3 \sqrt{3} - \frac{2 sin 40}{cos 20} + \frac{2 sin 20}{cos 80} - \frac{2 sin 100}{cos 40}$ $=3sqrt3-(2sin 40)/cos 20+(2sin 20)/cos 80-(2sin 100)/cos 40$

$= 3 \sqrt{3} - \frac{4 sin 20 cos 20}{cos 20} + \frac{4 sin 10 cos 10}{sin 10} - \frac{4 sin 40 cos 40}{cos 40}$ $=3sqrt3-(4sin 20cos 20)/cos 20+(4sin 10 cos 10)/sin 10-(4sin 40cos 40)/cos 40$

$= 3 \sqrt{3} - 4 sin 20 + 4 cos 10 - 4 sin 40$ $=3sqrt3-4sin 20+4cos 10-4sin 40$

$= 3 \sqrt{3} - 4 (sin 20 + sin 40) + 4 cos 10$ $=3sqrt3-4(sin 20+sin 40)+4cos 10$

$= 3 \sqrt{3} - 4 (2 sin 30 cos 1 0) + 4 cos 10$ $=3sqrt3-4(2 sin 30cos1 0)+4cos 10$

$= 3 \sqrt{3} - 4 (2 \cdot \frac{1}{2} \cdot cos 1 0) + 4 cos 10$ $=3sqrt3-4(2 *1/2*cos1 0)+4cos 10$

$= 3 \sqrt{3} - 4 cos 10 + 4 cos 10$ $=3sqrt3-4cos 10+4cos 10$

$= 3 \sqrt{3}$ $=3sqrt3$

Answer 4 · 2018-07-26T14:24:53

Explanation in below

Explanation:

$x = tan 20 + tan 80 + tan 140$ $x=tan20+tan80+tan140$

= $\frac{sin 20}{cos 20} + \frac{sin 80}{cos 80} + tan (180 - 40)$ $sin20/cos20+sin80/cos80+tan(180-40)$

= $\frac{cos 80 \cdot sin 20 + sin 80 \cdot cos 20}{cos 80 \cdot cos 20} - tan 40$ $(cos80*sin20+sin80*cos20)/(cos80*cos20)-tan40$

= $\frac{sin (80 + 20)}{cos 80 \cdot cos 20} - \frac{sin 40}{cos 40}$ $sin(80+20)/(cos80*cos20)-sin40/cos40$

= $\frac{sin 100}{cos 80 \cdot cos 20} - \frac{sin 40}{cos 40}$ $sin100/(cos80*cos20)-sin40/cos40$

= $\frac{sin 80}{cos 80 \cdot cos 20} - \frac{sin 40}{cos 40}$ $sin80/(cos80*cos20)-sin40/cos40$

= $(sin80*cos40-cos80*sin40*cos20)/(cos80*cos40*cos20)$

= $(sin20*(8sin80*cos40-8cos80*sin40*cos20))/(8cos80*cos40*cos20*sin20)$

= $(sin20*(4sin120+4sin40-4cos20*(sin120-sin40)))/(4cos80*cos40*sin40)$

= $(sin20*(4sin120+4sin40-4sin120*cos20+4sin40*cos20))/(2cos80*sin80)$

= $(sin20*(4sin60+4sin40-4sin60*cos20+4sin40*cos20))/(sin160)$

= $(sin20*(4sin60+4sin40-2sin80-2sin40+2sin60+2sin20))/(sin20)$

= $6sin60+2sin40-2sin80+2sin20$

= $3sqrt3+2sin20-(2sin80-2sin40)$

= $3sqrt3+2sin20-4cos60*sin20$

= $3sqrt3+2sin20-2sin20$

= $3sqrt3$

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