If tan θ + sec θ = √3, then the principal value of θ + \(\fra

1.	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» 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 + \(\frac{π}{6}\) = \(\frac{π}{3}\).

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»

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 + \(\frac{π}{6}\) = \(\frac{π}{3}\).