If tan 3θ+tan5θ+tan 3θ=1 , then θ is equal to

Question

If tan 3θ+tan5θ+tan 3θ=1 , then θ is equal to (A)  (nπ/8)+(π/16), n ∈ Z  (B)  (nπ/4)+(π/32), n ∈ Z  (C)  (nπ/8)+(π/32), n ∈ Z  (D)  (nπ/8)+(π/12), n ∈ Z

Tardigrade Admin · Accepted Answer

We have, tan 30 + tan 50 + tan 30 tan 50 = 1 ⇒ tan 30 + tan 50 = 1 - tan 30 tan 50 ⇒ (tan 3θ + tan 5θ/1-tan 3θ tan 5θ) = 1 ⇒ tan (3θ + 5θ) = 1 ⇒ tan 8θ = tan (π/4) ⇒ 2θ = n π + (π/4), n ∈ Z ⇒ θ = (nπ/8) + (π/32) , n∈ Z

Q. If tan $3 θ + t an 5 θ + t an 3 θ = 1$ , then $θ$ is equal to

$\frac{nπ}{8} + \frac{π}{16}, n \in Z$

$\frac{nπ}{4} + \frac{π}{32}, n \in Z$

$\frac{nπ}{8} + \frac{π}{32}, n \in Z$

$\frac{nπ}{8} + \frac{π}{12}, n \in Z$

Q. If tan 3θ+tan5θ+tan3θ=1 , then θ is equal to

8nπ​+16π​,n∈Z

4nπ​+32π​,n∈Z

8nπ​+32π​,n∈Z

8nπ​+12π​,n∈Z

We have, tan30+tan50+tan30tan50=1 ⇒tan30+tan50=1−tan30tan50 ⇒1−tan3θtan5θtan3θ+tan5θ​=1 ⇒tan(3θ+5θ)=1 ⇒tan8θ=tan4π​ ⇒2θ=nπ+4π​,n∈Z ⇒θ=8nπ​+32π​,n∈Z

Q. If tan $3 θ + t an 5 θ + t an 3 θ = 1$ , then $θ$ is equal to

$\frac{nπ}{8} + \frac{π}{16}, n \in Z$

$\frac{nπ}{4} + \frac{π}{32}, n \in Z$

$\frac{nπ}{8} + \frac{π}{32}, n \in Z$

$\frac{nπ}{8} + \frac{π}{12}, n \in Z$