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

4 Answers
maganbhai P.
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)#

P dilip_k
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#

P dilip_k
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#

Cem Sentin
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#

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