If tan θ+ tan 2 θ+√3 tan θ tan 2 θ=√3, then the general values of θ are

Question

If tan θ+ tan 2 θ+√3 tan θ tan 2 θ=√3, then the general values of θ are (A) (3 n+1) (π/3), n ∈ Z (B) (3 n+1) (π/9), n ∈ Z (C) (3 n+1) (π/6), n ∈ Z (D) (2 n+1) (π/9), n ∈ Z

Tardigrade Admin · Accepted Answer

It is given that tan θ+ tan 2 θ+√3 tan θ tan 2 θ=√3 ⇒ ( tan θ+ tan 2 θ/1- tan θ tan 2 θ)=√3= tan (π/3) ⇒ tan (3 θ)= tan (π/3) ⇒ 3 θ=n π+(π/3), n ∈ Z ⇒ θ=(3 n+1) (π/9), n ∈ Z

Q. If $tan θ + tan 2 θ + 3 tan θ tan 2 θ = 3$ , then the general values of $θ$ are

$(3 n + 1) \frac{π}{3}, n \in Z$

$(3 n + 1) \frac{π}{9}, n \in Z$

$(3 n + 1) \frac{π}{6}, n \in Z$

$(2 n + 1) \frac{π}{9}, n \in Z$

It is given that
$tan θ + tan 2 θ + 3 tan θ tan 2 θ = 3$
$\Rightarrow \frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = 3 = tan \frac{π}{3}$
$\Rightarrow tan (3 θ) = tan \frac{π}{3}$
$\Rightarrow 3 θ = nπ + \frac{π}{3}, n \in Z$
$\Rightarrow θ = (3 n + 1) \frac{π}{9}, n \in Z$

Q. If tanθ+tan2θ+3​tanθtan2θ=3​, then the general values of θ are

(3n+1)3π​,n∈Z

(3n+1)9π​,n∈Z

(3n+1)6π​,n∈Z

(2n+1)9π​,n∈Z