If tan θ + sec θ = √3 , then the principal value of θ + π/6 is equal to

Question

If tan θ + sec θ = √3, then the principal value of θ + \(\frac{π}{6}\) is equal to 

(a) \(\frac{π}{4}\)

(b) \(\frac{π}{3}\) 

(c) \(\frac{2π}{3}\) 

(d) \(\frac{3π}{4}\)

Answer 1

Answer: (b) = \(\frac{π}{3}\)

tan θ + sec θ = √3

⇒ \(\frac{sin\,\theta}{cos\,\theta}\) + \(\frac{1}{cos\,\theta}\) = √3 

⇒ sin θ + 1 = √3 cos θ

⇒ \(\frac{\sqrt{3}}{2}\) cos θ – \(\frac{1}{2}\) sin θ = \(\frac{1}{2}\) 

⇒ cos (θ + \(\frac{π}{6}\)) = cos\(\frac{π}{3}\)

⇒ Principal value of θ + \(\frac{π}{6}\) = \(\frac{π}{3}\).