If √3 cos θ – sin θ = 1, then θ is(a) π (b) π/2 (c) π/3 (

1.	If √3 cos θ – sin θ = 1, then θ is(a) π (b) π/2 (c) π/3 (d) π/6
Answer» Answer : (d) π/6 √3 cos θ – sin θ = 1 Dividing throughout by \(\sqrt{(\sqrt{3})^2+(-1)^2}\) = 2 We have \(\frac{\sqrt{3}}{2}\) cos θ - \(\frac{1}{2}\) sin θ = \(\frac{1}{2}\) ⇒ sin π/3 cos θ – cos π/3 sin θ = \(\frac{1}{2}\) ⇒ sin (π/3 - θ) = sin π/6 ⇒ π/3 - θ = π/6 ⇒ θ = π/6.

If √3 cos θ – sin θ = 1, then θ is(a) π (b) π/2 (c) π/3 (d) π/6

Answer»

Answer : (d) π/6

√3 cos θ – sin θ = 1

Dividing throughout by \(\sqrt{(\sqrt{3})^2+(-1)^2}\) = 2 We have

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

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

⇒ sin (π/3 - θ) = sin π/6

⇒ π/3 - θ = π/6

⇒ θ = π/6.