Prove that tan20+tan80+tan140=3sqrt3?

4 Answers
Jul 23, 2018

Please see below.

Explanation:

We take ,

LHS=tan 20^circ+tan80^circ+tan140^circ

color(white)(LHS)=tan20^circ+tan(60^circ+20^circ)+tan(120^circ+20^circ)

color(white)(LHS)=tan20^circ+(tan60^circ+tan20^circ)/(1-tan60^circtan20^circ)+(tan120^circ+tan20^circ)/(1-tan120^circtan20^circ)

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

LHS=t+(sqrt3+t)/(1-sqrt3t)+(-sqrt3+t)/(1+sqrt3t)

color(white)(LHS)=t+{(sqrt3+t)(1+sqrt3t)+(-sqrt3+t)(1-sqrt3t))/((1-sqrt3t)(1+sqrt3t))

color(white)(LHS)=t+(sqrt3+3t+t+sqrt3t^2-sqrt3+3t+t-sqrt3t^2)/(1-3t^2)

color(white)(LHS)=t+(8t)/(1-3t^2)

color(white)(LHS)=(t-3t^3+8t)/(1-3t^2)

color(white)(LHS)=(9t-3t^3)/(1-3t^2)

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

color(white)(LHS)=3[(3tan20^circ-tan^3 20^circ)/(1-3tan^2 20^circ)]

color(white)(LHS)=3[tan3(20^circ)]toApply(2) for theta=20^circ

LHS=3tan60^circ

LHS=3sqrt3=RHS

Note :

(1) tan(A+B)=(tanA+tanB)/(1-tanAtanB)

(2)tan3theta=(3tantheta-tan^3theta)/(1-3tan^2theta)

Jul 25, 2018

LHS=tan20+tan80+tan140

=tan20+tan80+tan(180-40)

=tan20+tan80-tan 40

=tan20+sin 80/cos 80-sin 40/cos 40

=sin 20/cos 20+(sin 80cos 40-cos 80sin 40)/(cos 80cos 40)

=(sin 20cos 80cos 40+sin 40cos 20) /(cos 20cos 80cos 40)

Now denominator of this expression

=cos 20cos 80cos 40

=(4*2sin 20cos 20cos 40cos 80)/(8sin 20)

=(2*2sin 40cos 40cos 80)/(8sin 20)

=(2sin 80cos 80)/(8sin 20)

=(sin 160)/(8sin 20)

=(sin (180-20))/(8sin 20)

=(sin 20)/(8sin 20)

=1/8

Hence

LHS=8(sin 20cos 80cos 40+sin 40cos 20)

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

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

=4sin 20(-1/2+cos 40)+4(sqrt3/2+sin 20)

=-2sin 20+4sin 20cos 40+2sqrt3+4sin 20

=4sin 20cos 40+2sqrt3+2sin 20

=2(sin 60-sin 20)+2sqrt3+2sin 20

=2(sqrt3/2-sin 20)+2sqrt3+2sin 20

= sqrt3-2sin 20+2sqrt3+2sin 20

=3sqrt3

Jul 25, 2018

A funny approach utilising the anwer 3sqrt3 given.

We can write LHS as follows as we know sqrt3=tan 60

LHS=tan 20 +tan 80 + tan 140

=3sqrt3+(tan 20-tan 60)+(tan 80-tan 60) +(tan 140-tan 60)

=3sqrt3+(tan 20-tan 60)+(tan 80-tan 60) +(tan (180-40)-tan 60)

=3sqrt3+(tan 20-tan 60)+(tan 80-tan 60) -(tan 40+tan 60)

=3sqrt3+(sin 20/cos 20-sin 60/cos60)+(sin 80/cos 80-sin 60/cos60 )-(sin 40/cos40+sin 60/cos60)

=3sqrt3-sin(60- 20)/(cos 20cos60)+sin (80-60)/(cos 80cos60 )-sin (60+40)/(cos40cos60)

=3sqrt3-(2sin 40)/cos 20+(2sin 20)/cos 80-(2sin 100)/cos 40

=3sqrt3-(4sin 20cos 20)/cos 20+(4sin 10 cos 10)/sin 10-(4sin 40cos 40)/cos 40

=3sqrt3-4sin 20+4cos 10-4sin 40

=3sqrt3-4(sin 20+sin 40)+4cos 10

=3sqrt3-4(2 sin 30cos1 0)+4cos 10

=3sqrt3-4(2 *1/2*cos1 0)+4cos 10

=3sqrt3-4cos 10+4cos 10

=3sqrt3

Jul 26, 2018

Explanation in below

Explanation:

x=tan20+tan80+tan140

=sin20/cos20+sin80/cos80+tan(180-40)

=(cos80*sin20+sin80*cos20)/(cos80*cos20)-tan40

=sin(80+20)/(cos80*cos20)-sin40/cos40

=sin100/(cos80*cos20)-sin40/cos40

=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