39The pi/2-theta conversion part....I think it works only for first quadrant angles. Here we are considering all four quadrants.
Do correct me if I'm wrong.
ques- solve for θ:
tan2θ.tanθ = 1
i found two ways of solving this but both of them are giving different answers
I - by using tan2θ = 2tanθ1-tan2θ
and solving for tanθ ,i got
tanθ = ±1√3
ie θ = nΠ± Π6
II - tan2θ = cotθ
tan2θ = tan(Π2 - θ)
2θ = nΠ+Π2 - θ
ie θ = Π3(n + 12)
both the answers are coming out to be different .....
plz someone tell where is my mistake????
I know 2nd method is wrong but plz explain how
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11 Answers
39
23that conversion is ok , it is valid for all angles
i think where u r making mistake is : taking both the natural nos as n . Take one as m and other as n. there isnt any reason for them to be equal , they may be related by some other equation.
1yes qwerty that is right ..
let the answer be:
θ = Π3(m + 12)
but the problem is that both answers give different set of angles ,for example taking
m = 1 (in second solution) gives θ = Π2
whereas the first solution's answer can never yield θ = Π2 for any value of 'n'
341Perhaps it would have been simpler to just use
\tan \theta \tan 2\theta =1 \Rightarrow \tan 3\theta=0
1yes euclid
tan3θ = ∞
using this the solution would be same as second one
θ = Π3(x + 12) .......where x is any integer
but the answer given is, θ = nΠ± Π6 same as the first solution
1There is a small mistake that can happen .
tan θ x tan 2 θ = 1 may imply that ;
tan 3 θ = ∞ ,
if and only if , tan θ + tan 2 θ ≠0 ,
But in this case , can this happen ?
No , this can't happen as tan θ and tan 2 θ must have the same sign .
So the arguments hold .
1thanx ricky
i think i got my mistake
the set of angles, θ = nΠ± Π6 is a subset of
the set θ = Π3(m + 12)
the extra values of θ that second set contains are = Π2 , 5Π2 , 9Π2 .........
what i did in solution II can't be used because converting
tan2θtanθ=1 to tan2θ = cotθ would be like
tan2θcotθ = 1
and at Î 2 , 5Î 2 , 9Î 2 .........
tan2θcotθ = 00 ....... an undefined term
so we can't bring cot in Dr
but still im not able to find why tan3θ = ∞ can't be used