If θ + φ = (π/6), then (√3 + tan θ)( √3 + tan φ) =

Question

If θ + φ = (π/6), then (√3 + tan θ)( √3 + tan φ) =  (A) 1 (B) -1 (C) 4 (D) -4

Tardigrade Admin · Accepted Answer

θ+φ = (π/6) ⇒ tan(θ +φ) = tan (π/6) ⇒ ( tan θ + tanφ/1- tanθ tan φ) = (1/√3) ⇒ √3 ( tanθ + tan φ) = 1 - tan θ tanφ ⇒ tanθ tan φ + √3 tanθ +√3 tanφ = 1 Add 3 to both sides and factorise L.H.S.

Q. If $θ + ϕ = \frac{π}{6}$ , then $(3 + tan θ) (3 + tan ϕ) =$

1

-1

4

-4

$θ + ϕ = \frac{π}{6} \Rightarrow tan (θ + ϕ) = tan \frac{π}{6}$
$\Rightarrow \frac{t a n θ + t a n ϕ}{1 - t a n θ t a n ϕ} = \frac{1}{3}$
$\Rightarrow 3 (tan θ + tan ϕ) = 1 - tan θ tan ϕ$
$\Rightarrow tan θ tan ϕ + 3 tan θ + 3 tan ϕ = 1$
Add 3 to both sides and factorise L.H.S.

Q. If θ+ϕ=6π​, then (3​+tanθ)(3​+tanϕ)=

1

-1

4

-4

θ+ϕ=6π​⇒tan(θ+ϕ)=tan6π​ ⇒1−tanθtanϕtanθ+tanϕ​=3​1​ ⇒3​(tanθ+tanϕ)=1−tanθtanϕ ⇒tanθtanϕ+3​tanθ+3​tanϕ=1 Add 3 to both sides and factorise L.H.S.

Q. If $θ + ϕ = \frac{π}{6}$ , then $(3 + tan θ) (3 + tan ϕ) =$

$θ + ϕ = \frac{π}{6} \Rightarrow tan (θ + ϕ) = tan \frac{π}{6}$
$\Rightarrow \frac{t a n θ + t a n ϕ}{1 - t a n θ t a n ϕ} = \frac{1}{3}$
$\Rightarrow 3 (tan θ + tan ϕ) = 1 - tan θ tan ϕ$
$\Rightarrow tan θ tan ϕ + 3 tan θ + 3 tan ϕ = 1$
Add 3 to both sides and factorise L.H.S.