If $3(\cot^2 \theta - \cos^2 \theta) = 1 - \sin^2 \theta, 0^\circ < \theta < 90^\circ$, then the $\theta$ equal to:

30° 60° 45° 15°

60°

3( cot²θ - cos²θ ) = 1 - sin²θ { using , sin²θ + cos²θ = 1 } 3cot²θ - 3cos²θ  = cos²θ 3cot²θ = 4cos²θ 3cosec²θ = 4 cosec²θ = \(\frac{4}{3}\) cosecθ = \(\frac{2}{√3}\) { we know, cosec60º = \(\frac{2}{√3}\) } So,  θ= 60º