1bhaiyya cud u plzzzzzzzz explain............y did u take n1 for He as 2 and n2 as 4?????
(n2=4 i kno u hav assumed it to make me understand but y n1 as 2)??
This is a fun problem. May i request that only the students attempt this one?
If a,b and c are all distinct real numbers, then find values of a,b and c such that
\frac{a-b}{1+ab} + \frac{b-c}{1+bc} + \frac{c-a}{1+ca} = 0
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21 Answers
62no i was just showing that you have to go from z=1 to z=2
so if you simply multiplied and divided by 4.. then the form would look like what it does here.....
now suppose the queston was
H-Li2+ mixture
Li is excited to the 2nd state. Then what would the answer be?
162R.12(1/12-1/22)
=
R22(1/22-1/42)
does this finally give you the correct answer?
1ohhhhhhhh cool................................THANKS FOR ANOTHER TRICK..........................
62thi sis no twhat i meant ;)
well this surprising result:
tan A. tan 2A tan 3A= Tan 3A- Tan A - Tan 2A
1tan2A=2tanA/1-tan2A
tan3A=tan(A+2A)=(tanA+tan2A)/(1-tanAtan2A)
=......expand tan2A here..............
but bhaiyya this is a normal formula naa...........
every1 knos these........
62Akand..
this is a very old problem
find the relation ship between
tan A, tan 2A and tan 3A
1wow prophet sir...........tht method wud be awesum..........thnx for dis trick......ill remember it.........
341@Akand One more useful fact is that
\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1- \tan A \tan B - \tan B \tan C - \tan C \tan A}
Here A+B+C = 0 and hence it follows that tan A + tan B + tan C = tan A tan B tan C
[you would have encountered a similar argument when A,B C are angles of a triangle]
See Loney for a general formula for
\tan(\theta_1 + \theta_2 +...+\theta_n)
1yup bhaiyya..........................but who cud think like tht except rohan.........
k thnx anyways......nice trick though........TargetIIT is helpin me a lot....luv u guysssssss
1is there a formula like that??????????????????????????
OMG.....................JEE is comin closer and im not prepared yet.........
I swear bhaiyya...............who cud think like that
tan(a+b)=tana+tanb/1-tanatanb
then cross multiplying and using here.............uff tooo much..........
waaaaahhhhhhhhhhhhhhhhhh..............
62first step is
tan (a) + tan b= tan (a+b)(1-tana tan b)
where a=x-y
b=y-z
1rohan...............thnx a lot man............i was tryin this from tht time...........but i still cudnt get ur first step..........
1after akands step
just put
tan(x-y) + tan(y-z)=tan(x-z)(1-tan(x-y)tan(y-z))
it reduces to
tan(x-y)-tan(x-y)tan(x-z)tan(y-z)+tan(z-x)
=
tan(z-x)tan(x-y)tan(y-z)=0
either
tan(x-y)=0 or tan(y-z)=0 or tan(z-x)=0
from here we vget all solutions
1wel i think we can take a as tanx b as tany and c as tanz
so it reduces to
tan(x-y)+tan(y-z)+tan(z-x)=0