Question

a , b, c are 3 consecutive terms of an A.P if tan a , tan b , tan c (b= not equal pie/2) are also in A.P then?


(a) tan b = 2 tan a

(b) tan a = tan b = tan c

Answer 1

@kartiksriramk @dumbcow @hartnn

Answer 2

are these 2 options? or to be proved??

Answer 3

options

Answer 4

nop.....The answer should be (b) tan a = tan b = tan c

Answer 5

hmmm,working on it,wait

Answer 6

@mahmit2012 @mahmit2012 @mahmit2012 @mahmit2012

Answer 7

i got it but its a very lengthy solution.....
u know that if 3 numbers are in AP and also in GP then these three are equal??
because i am getting tanA,tanB and tanC in GP
that is i am getting tanA*tanC=tanB^2=3

Answer 8

there is another lengthy solution
use fact that : b-a = c-b --> tan(b-a) = tan(c-b)
also because tan is in AP --> tan(b) = (tan a +tan c)/2

now use difference formula for tangent and substitute in for tan(b)
this will prove that tan(a) = tan(c)

Answer 9

a+c=2b----->'X'
tan a +tan c =2 tan b------>'Y'
A well known result :
tana*tanb*tanc=tana+tanb+tanc----->'Z'
from 'Y' and 'Z',
tan a tan b tan c-tan b=2 tan b
or tan a*tan c= 3-------->'W'

now from 'X',tan (a+c)=tan 2b
\[\frac{\tan a + \tan c}{1- \tan a*\tan c}=\frac{2 \tan b}{1-(\tan b)^2}\]
\[\frac{2 \tan b }{1-3}=\frac{2 \tan b}{1- (\tan b)^2}\]
from 'W' and 'X'

which gives (tan b)^2=3

implies,tan a* tanc = (tan b)^2
implies tan a ,tan b,tan c are also in GP and in AP also
hence, tana=tanb=tanc

@Yahoo