If Tanθ+Tan2θ+3 Tanθ =3 then θ=

Tanθ+Tan2θ+3TanθTan2θ=3Tanθ+Tan2θ1−TanθTan2θ=3⇒Tan3θ=3⇒3θ=nπ+π3θ=nπ3+π9,n∈z

If Tanθ+Tan2θ+3 Tanθ =3 then θ=

Tanθ+Tan2θ+3TanθTan2θ=3Tanθ+Tan2θ1−TanθTan2θ=3⇒Tan3θ=3⇒3θ=nπ+π3θ=nπ3+π9,n∈z

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $T a n θ + T a n 2 θ + \sqrt{3} T a n θ = \sqrt{3}$ then $θ =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{n π}{3} + \frac{π}{9}, n \in z$

$\begin{array}{l} \frac{n π}{3} + \frac{π}{12}, n \in z \end{array}$

$n π + {(- 1)}^{n} \frac{π}{6}, n \in z$

$\frac{n π}{3} + \frac{π}{4}, n \in z$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} T a n θ + T a n 2 θ + \sqrt{3} T a n θ T a n 2 θ = \sqrt{3} \\ \frac{T a n θ + T a n 2 θ}{1 - T a n θ T a n 2 θ} = \sqrt{3} \Rightarrow T a n 3 θ = \sqrt{3} \Rightarrow 3 θ = n π + \frac{π}{3} \\ θ = \frac{n π}{3} + \frac{π}{9}, n \in z \end{array}$