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