If $$\frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3, 0^\circ < \theta < 90^\circ$$, then the value of $$\cot \theta + \cosec \theta$$ is:

Given that,

$$\frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3$$

$$\Rightarrow \cos^2 \theta=3( \cot^2\theta-\cos^2 \theta)$$

$$\Rightarrow \cos^2 \theta +3 \cos^2 \theta=3\cot^2\theta$$

$$\Rightarrow 4\cos^2 \theta =3\cot^2\theta$$

$$\Rightarrow 4\cos^2 \theta =3\dfrac{\cos^2 \theta}{\sin^2 \theta}$$

$$\Rightarrow \sin^2 \theta=\dfrac{3}{4}$$

$$\Rightarrow \sin^2 \theta=\dfrac{3}{4}$$

$$\Rightarrow \sin \theta=\dfrac{\sqrt 3}{2}=\sin 60$$

$$\theta=60$$

Now, substituting the values,

$$\Rightarrow \cot \theta + \cosec \theta =\cot 60^\circ+\cosec 60^\circ$$

$$\Rightarrow \cot 60^\circ+\cosec 60^\circ=\dfrac{1}{\sqrt 3}+\dfrac{2}{\sqrt 3}=\dfrac{3}{\sqrt3}=\sqrt{3}$$