1hey yes.. btw i have another question pls answer that as well...
thanks for this question :)
prove tan a + tan 2a + tan 3a = tan a . tan 2a . tan 3a
my friend gave me from school.. i cant get anything to work on.. this is so tough :(
-
UP 0 DOWN 0 0 8
8 Answers
62hmm.. this is not as tough my friend...
u need to use
tan 3a = tan (a+2a) and expand...
I hope u can work the rest ur self.. just write it on paper.. it is very simple...
But if i remember, in my school days, this question defeated me as well.. i think my maths teacher in shool asked this one :)
Probably from RS aggrawal if i remember correctly
1but one thing.. ur answer was very fast.. i just give the question and in 2 minutes ur answer is there!! good man...
i think u can work some on the tests... i want longer tests as well...
1
1This is an easy question.......as nishant stated use tan(a+2a)
tan 3a=(tan a + tan 2a )/1-tan a tan 2a
now cross multiply
you get tan3a - tan a tan2a tan3a
But I think final ans will be tan3a -tan a -tan2a=tana.tan2a.tan3a
Nishant pls cross check & tell me...
1the identity
tana+tan2a+tan3a = tana.tan2a.tan3a
will hold only for a = ±1,0
1You can use the general formula for addition of n-terms of tan
tan ( a + b + c+ .....) = S1 - S3 + S4 -...
1- S2 + S4 - ....
where S1 = tan a + tan b + tanc +......
S2 = tan a * tan b + tan a * tanc +..... (sum of tan's taken to at a time)
Here, tan a + tan 2a + tan 3a = tan(-a) + tan (-2a) + tan 3a
Hence tan (3a - 2a -a ) = 0
Therefore , S1 = S3
Hence the result