Answer: C
\[ 3x+ \cfrac{3}{x} = \csc \alpha + \cot \alpha \] or, \[ 3\left(x+\cfrac{1}{x}\right) = \csc \alpha + \cot \alpha \] Similarly, \[ 3x - \cfrac{3}{x} = \csc \alpha - \cot \alpha \] or, \[ 3(x - \cfrac{1}{x}) = \csc \alpha - \cot \alpha \] ∴ \[ 3(x+\cfrac{1}{x}) \times 3(x-\cfrac{1}{x}) \] \[ = (\csc \alpha + \cot \alpha)(\csc \alpha - \cot \alpha) \] or, \[ 3 \times 3 \left(x^2 - \cfrac{1}{x^2} \right) = \csc^2 \alpha - \cot^2 \alpha \] \[ 3 \times 3 \left(x^2 - \cfrac{1}{x^2} \right) = 1 \] \[ 3 \left(x^2 - \cfrac{1}{x^2} \right) = \cfrac{1}{3} \]
\[ 3x+ \cfrac{3}{x} = \csc \alpha + \cot \alpha \] or, \[ 3\left(x+\cfrac{1}{x}\right) = \csc \alpha + \cot \alpha \] Similarly, \[ 3x - \cfrac{3}{x} = \csc \alpha - \cot \alpha \] or, \[ 3(x - \cfrac{1}{x}) = \csc \alpha - \cot \alpha \] ∴ \[ 3(x+\cfrac{1}{x}) \times 3(x-\cfrac{1}{x}) \] \[ = (\csc \alpha + \cot \alpha)(\csc \alpha - \cot \alpha) \] or, \[ 3 \times 3 \left(x^2 - \cfrac{1}{x^2} \right) = \csc^2 \alpha - \cot^2 \alpha \] \[ 3 \times 3 \left(x^2 - \cfrac{1}{x^2} \right) = 1 \] \[ 3 \left(x^2 - \cfrac{1}{x^2} \right) = \cfrac{1}{3} \]