Question

If tan⁡3θ−1tan⁡3θ+1=3, then the general value of θ is:

• A. nπ3−π12,n∈Z
• B. nπ+7π12,n∈Z
• C. nπ3+7π36,n∈Z
• D. nπ+π12,n∈Z

Answer 1

Answer: (C)

nπ3+7π36,n∈Z

Answer 2

Explanation:
Given: tan⁡3θ−1tan⁡3θ+1=3
We have to find the general value of θ.
Consider, tan⁡3θ−1tan⁡3θ+1=3⇒tan⁡3θ−tan⁡π41+tan⁡3θ⋅tan⁡π4=3[Using trigonometric ratios ]
⇒tan⁡(3θ−π4)=tan⁡π3[Using trigonometric formula of tangent function ]⇒3θ−π4=nπ+π3,n∈Z[Using general solution of tangent function]⇒3θ=nπ+π3+π4,n∈Z⇒3θ=nπ+4π+3π12,n∈Z⇒θ=nπ3+7π36,n∈Z
Hence, the correct option is (C).