General solution of the equation tan θ tan 2 θ=1 is given by

Question

General solution of the equation tan θ tan 2 θ=1 is given by (A) (2 n+1) (π/4), n ∈ I (B) n π+(π/6), n ∈ I (C) n π-(π/6), n ∈ I (D) n π ± (π/6), n ∈ I

Tardigrade Admin · Accepted Answer

tan θ tan 2 θ=1 ∴ tan θ (2 tan θ/1- tan 2 θ)=1 ∴ 2 tan 2 θ=1- tan 2 θ ∴ 3 tan 2 θ=1 ∴ θ= n π ± (π/6)

Q. General solution of the equation $tan θ tan 2 θ = 1$ is given by

$(2 n + 1) \frac{π}{4}, n \in I$

$nπ + \frac{π}{6}, n \in I$

$nπ - \frac{π}{6}, n \in I$

$nπ \pm \frac{π}{6}, n \in I$

$tan θ tan 2 θ = 1$
$∴ tan θ \frac{2 t a n θ}{1 - t a n ^{2} θ} = 1$
$∴ 2 tan^{2} θ = 1 - tan^{2} θ$
$∴ 3 tan^{2} θ = 1$
$∴ θ = nπ \pm \frac{π}{6}$

Q. General solution of the equation tanθtan2θ=1 is given by

(2n+1)4π​,n∈I

nπ+6π​,n∈I

nπ−6π​,n∈I

nπ±6π​,n∈I