trigo doubt

The 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.

that 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.

yes 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'

sir, i m getting 'tan3θ = ∞'... how can it be zero? plzz explain

yes 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

There 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 .

thanx 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 D^r

but still im not able to find why tan3θ = ∞ can't be used