1+tan@+tan^2@+tan^3@+....♾️= (√3+1)/2
where 0<@<π/4
consider @ as tthetha
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Answers
Step-by-step explanation:
Given:-
1 + Tan θ + Tan²θ + Tan³θ +..∞ = (√3+1)/2
Where 0 < θ ≤ π/4
To find :-
Find the value of θ ?
Solution :-
Given that
1 + Tan θ + Tan²θ + Tan³θ +..∞ = (√3+1)/2
Where 0 < θ ≤ π/4
First term (a) = 1
Common ratio (r) = Tan θ/1 = Tan θ
=>r = Tan² θ/ Tan θ = Tan θ
Since the common ratio is same throughout the series
Given series is in the G.P.
We know that
The sum of the ∞ terms in the GP = a/ (1-r)
=> S ∞ = a / (1-r)
=> S ∞ = 1/ (1- Tan θ)
We have
=> 1/ (1- Tan θ) = (√3+1)/2
On applying cross multiplication then
=> (√3+1)(1- Tan θ) = 2×1
=> (√3+1)(1 - Tan θ) = 2
=> √3-√3 Tan θ + 1 - Tan θ = 2
=> -√3 Tan θ - Tan θ = 2 - √3 -1
=> -√3 Tan θ - Tan θ = 1-√3
=> - Tan θ ( √3+1) = 1-√3
=> (√3+1) Tan θ = -(1-√3)
=> (√3+1) Tan θ = √3-1
=> Tan θ = (√3-1)/(√3+1)
=>Tan θ=(Tan60°-Tan45°)/(Tan 60°+ Tan 45°)
It can be written as
Tan θ =(Tan 60°-Tan45°)/(1+Tan60°Tan 45°)
It is in the form of
(TanA-TanB)/(1+TanA TanB)
We know that
Tan (A-B) = (Tan A -Tan B /(1+ Tan A Tan B)
Now , we have
Tan θ = Tan (60°-45°)
=> Tan θ = Tan 15°
=> θ = 15° or
=> θ = 15°×π/180°
=> θ = π/12
Therefore ,θ = 15° or π/12
Answer:-
The value of θ for the given problem is 15° (or) π/12
Used formulae:-
→ The sum of the ∞ terms in the GP
= a/ (1-r)
- a = First term
- r = Common ratio
→ Tan(A-B) = (Tan A -Tan B)/(1+Tan ATan B)
→ Tan 60° = √3
→ Tan 45° = 1
→ π = 180°