The general solution of cot θ + tan θ = 2 is1. \(\theta = n\pi + {

1.	The general solution of cot θ + tan θ = 2 is1. \(\theta = n\pi + {( - 1)^n}\frac{\pi }{8}\)2. \(\theta = \frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{6}\)3. \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\)4. \(\theta = \frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{8}\)
Answer» Correct Answer - Option 3 : \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\) Calculation: Given cot θ + tan θ = 2 \(\rm {1\over \tanθ} +\tan θ = 2\) tan² θ - 2 tan θ + 1 = 0 (tan θ - 1)² = 0 tan θ = 1 θ = \({\pi\over4}, {5\pi\over4}, {9\pi\over4}...\) = \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\), (n ∈ N)

The general solution of cot θ + tan θ = 2 is1. \(\theta = n\pi + {( - 1)^n}\frac{\pi }{8}\)2. \(\theta = \frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{6}\)3. \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\)4. \(\theta = \frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{8}\)

Answer» Correct Answer - Option 3 : \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\)

Calculation:

Given cot θ + tan θ = 2

\(\rm {1\over \tanθ} +\tan θ = 2\)

tan² θ - 2 tan θ + 1 = 0

(tan θ - 1)² = 0

tan θ = 1

θ = \({\pi\over4}, {5\pi\over4}, {9\pi\over4}...\) = \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\), (n ∈ N)