If $$12 \cot^2 \theta  - 31 \cosec \theta + 32 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$\tan \theta$$ will be:

As per the given question,

$$12 \cot^2 \theta - 31 \cosec \theta + 32 = 0, 0^\circ < \theta < 90^\circ$$

$$12(\cosec^2 \theta-1)-31 \cosec \theta+32=0$$

$$12 \cosec^2 \theta-31 \cosec \theta+20=0$$

Let $$\cosec \theta=x$$

$$12 x^2-31x+20=0$$

$$12x^2-15x-16x+20=0$$

$$3x(4x-5)-4(4x-5)=0$$

$$(4x-5)(3x-4)=0$$

So, $$x=\dfrac{5}{4}$$ or $$x=\dfrac{4}{3}$$

So, $$\sin \theta=\dfrac{4}{5}$$ or $$\sin \theta=\dfrac{3}{4}$$

Hence, $$\tan \theta=\dfrac{4}{3} or \tan \theta=\dfrac{3}{\sqrt 7}$$

$$\tan \theta=\dfrac{4}{3} or \tan \theta=\dfrac{3\sqrt 7}{7}$$