If $$\cot \theta = \sqrt{6}$$, then the value of $$\frac{\cosec^2 \theta + \sec^2 \theta}{\cosec^2 \theta - \sec^2 \theta}$$ is:

given 

$$\cot\theta = \sqrt{6}$$

now squaring both side we get

$$\cot^{2}\theta$$ =6

we also know $$\cot^{2}\theta = \frac{\cos^{2}\theta}{\sin^{2}\theta}$$

$$\cos^{2}\theta = 6\sin^{2}\theta$$

now adding both side 1 we get

$$\cot^{2}\theta$$ + 1 = 6+1 i.e 7

therefore $$\cosec^{2}\theta$$ = 7

now 

$$\frac{\cosec^{2}\theta + \sec^{2}\theta}{\cosec^{2}\theta - \sec^{2}\theta}$$

$$\frac{\frac{1}{\sin^{2}\theta} + \frac{1}{cos^{2}\theta}{\frac{1}{\sin^{2}\theta} - \frac{1}{cos^{2}\theta}$$

$$\frac{\cos^{2}\theta + \sin^{2}\theta}{\cos^{2}\theta - \sin^{2}\theta}$$

$$\frac{1}{\cos^{2}\theta - \sin^{2}\theta}$$ now applying the value we get

$$\frac{1}{6\sin^{2}\theta - \sin^{2}\theta}$$

$$\frac{1}{5\sin^{2}\theta}$$

$$\frac{1}{5}\cosec^{2}\theta$$ 

therefore $$\frac{7}{5}$$ is the answer