Let S= θ ∈(0, (π/2)): displaystyle∑m=19 sec (θ+(m-1) (π/6)) sec (θ+(m π/6))=-(8/√3) Then

Question

Let S= θ ∈(0, (π/2)): displaystyle∑m=19 sec (θ+(m-1) (π/6)) sec (θ+(m π/6))=-(8/√3) Then (A) S= (π/12)  (B) S= (2 π/3)  (C)  displaystyle∑θ ∈ S θ=(π/2) (D)  displaystyle∑θ ∈ S θ=(3 π/4)

Tardigrade Admin · Accepted Answer

Let α=θ+(m-1) (π/6) β=θ+m (π/6) So, β-α=(π/6) Here, displaystyle∑ m =19 sec α ⋅ sec β=∑ m =19 (1/ cos α ⋅ cos β) =2 displaystyle∑ m =19 ( sin (β-α)/ cos α ⋅ cos β)=2 displaystyle∑ m =19( tan β- tan α) =2 displaystyle∑ m =19( tan (θ+ m (π/6))- tan (θ+( m -1) (π/6))) =2( tan (θ+(9 π/6))- tan θ)=2(- cot θ- tan θ)=-(8/√3) (Given) ∴ tan θ+ cot θ=(4/√3) ⇒ tan θ=(1/√3) text or √3 So, S= (π/6), (π/3) displaystyle∑θ ∈ S θ=(π/6)+(π/3)=(π/2)

Q. Let $S = {θ \in (0, \frac{π}{2}) : m = 1 \sum 9 sec (θ + (m - 1) \frac{π}{6}) sec (θ + \frac{mπ}{6}) = - \frac{8}{3}}$ Then

$S = {\frac{π}{12}}$

$S = {\frac{2 π}{3}}$

$θ \in S \sum θ = \frac{π}{2}$

$θ \in S \sum θ = \frac{3 π}{4}$

Q. Let S={θ∈(0,2π​):m=1∑9​sec(θ+(m−1)6π​)sec(θ+6mπ​)=−3​8​} Then

S={12π​}

S={32π​}

θ∈S∑​θ=2π​

θ∈S∑​θ=43π​

Q. Let $S = {θ \in (0, \frac{π}{2}) : m = 1 \sum 9 sec (θ + (m - 1) \frac{π}{6}) sec (θ + \frac{mπ}{6}) = - \frac{8}{3}}$ Then

$S = {\frac{π}{12}}$

$S = {\frac{2 π}{3}}$

$θ \in S \sum θ = \frac{π}{2}$

$θ \in S \sum θ = \frac{3 π}{4}$