Question

Prove That :
___________
Tan20Tan30Tan40=Tan10

Answer 1

@experimentX : I really appreciate your effort for helpin :) ,TYSM bro ..and if u got tired just leave it ,i'am sure i will find its solution soon ^_^

Answer 2

tan(20 + 30 + 40) = ?? is there formula for tan(A + B + C) ??

Answer 3

nope ,but we can say tan20=sin20/cos20 ,and tan40 =sin40/cos40

Answer 4

Btw @experimentX Its Tan(a*B*c) not Tan(A+B+C)

Answer 5

let's try it with sines and cosines
(sin 20 sin 30 sin 40)/(cos 20 cos 30 cos 40)
(sin 20 (cos 10 - cos 70))/(cos 20 (cos 10 + cos 70)
(sin 20 (cos 10 - sin 20))/(cos 20 (cos 10 + cos 20))
(sin 20 cos 10 - sin 20 sin 20 )/(cos 20 cos 10 + cos 20 cos 20)

Answer 6

maybe it might help to write it as (tan20tan40) / sqrt3 - just a thought..

Answer 7

this is going to quite long procedure
question is more like
http://www.wolframalpha.com/input/?i=Tan20Tan30tan40tan60tan80+%3D+sqrt%283%29

Answer 8

http://www.goiit.com/posts/list/trignometry-find-the-value-of-tan20tan40tan60tan80-953230.htm;jsessionid=0CE4E927F29F2B0EE17B5692CDE1D8CC.node1#1071444

Answer 9

I think joeywhite might be on to something, since tan(30) has a well-known value it might be worth trying to re-write the expression in terms of tan(30)

Answer 10

\[\tan(20)\tan(30)\tan(40)=\tan(30-10)\tan(30)\tan(30+10)\]

Answer 11

Oo .. yeah right
20 = 30 -10
40 = 30 +10

Answer 12

that leads to:\[\tan(30-10)\tan(30)\tan(30+10)=\frac{\tan(30)-\tan(10)}{1+\tan(30)\tan(10)}*\tan(30)*\frac{\tan(30)+\tan(10)}{1-\tan(30)\tan(10)}\]\[\qquad=\frac{\tan^2(30)-\tan^2(10)}{1-\tan^2(30)\tan^2(10)}*\tan(30)\]\[\qquad=\frac{1-3\tan^2(10)}{3-\tan^2(10)}*\tan(30)\]

Answer 13

(sin 30 cos 10 - cos 30 sin 10) sin 30 (sin 30 cos 10 + cos 30 sin 10)
-------------------------------------------------------------------------------------------
(cos 10 cos 30 + sin 10 sin 30) cos 30 (cos 10 cos 30 - sin 10 sin 30)

1/4 ( cos 10 - sqrt(3) sin 10) sin 30 ( cos 10 + sqrt(3) sin 10)
-------------------------------------------------------------------------------------------
1/4 (cos 10 sqrt(3) + sin 10 ) cos 30 (cos 10 sqrt(3) - sin 10 )

( cos^2 10 - 3 sin^2 10) sin 30
-----------------------------------------
(3 cos^2 10 - sin^2 10 ) cos 30

Answer 14

we can then use the triple angle formula for tan(30) to get:\[\tan(30)=\frac{3\tan(10)-\tan^3(10)}{1-3\tan^2(10)}\]

Answer 15

therefore:\[\tan(30)=\tan(10)*\frac{3-\tan^2(10)}{1-3\tan^2(10)}\]and this can be substituted into the result above to get the proof.

Answer 16

hope it makes sense Eyad?

Answer 17

Yea thats help @asnaseer ,But i discovered a simpler solution ..
Ty anyway :)

Answer 18

I'd be interested in the simpler solution if you have time to post it

Answer 19

@asnaseer :i'am totally sry about yesterday or should i say about last month cuz your reply was one month ago xD ,the internet connection went out :( ...Well the answer is not simple as u thing :D,i mean its not very simple ,anyway i wrote it and i will share upload them now to the web and post their link to u ..

Answer 20

one month ago?

Answer 21

which reply are you talking about?

Answer 22

hang on - do you mean its May where ever you are right now? Because here its still April.

Answer 23

I'm in the UK and its currently 9:18pm April 30th

Answer 24

oh,yea its may here :)..and thats mean one month ago xD

Answer 25

ok - that makes sense now - you confuzzled me for a second :D

Answer 26

and no need to apologise - I was just interested in a simpler solution - thats all

Answer 27

here is the answer : paper number 1