If $$2 \sin^2 \theta + 5 \cos \theta - 4 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$\cot \theta + \cosec \theta$$ is:

$$2 \sin^2 \theta + 5 \cos \theta - 4 = 0$$ 

$$2 \sin^2 \theta + 5 \cos \theta  = 4$$

Where $$0^\circ < \theta < 90^\circ$$

As per the above given limit of $$\theta$$, it can be $$30^\circ , 45^\circ and 60^\circ$$

When $$\theta = 30^\circ$$

$$2 \sin^2 30^\circ + 5 \cos 30^\circ = 4$$

$$2\left(\frac{1}{2}\right)^2+5\times\frac{\sqrt{\ 3}}{2}=4$$

This equation is not satisfied. So the value of $$\theta = 30^\circ$$ will not be possible.

When $$\theta = 45^\circ$$

$$2 \sin^2 45^\circ + 5 \cos 45^\circ = 4$$

$$2\left(\frac{1}{\sqrt{\ 2}}\right)^2+5\times\ \left(\frac{1}{\sqrt{\ 2}}\right)=4$$
This equation is not satisfied. So the value of $$\theta = 45^\circ$$ will not be possible.

When $$\theta = 60^\circ$$

$$2 \sin^2 60^\circ + 5 \cos 60^\circ = 4$$

$$2\times\ \left(\frac{\sqrt{\ 3}}{2}\right)^2+5\times\frac{1}{2}=4$$

$$\frac{3}{2}+\frac{5}{2}=4$$

$$\frac{8}{2}=4$$

4 = 4
This equation is satisfied. So the value of $$\theta = 60^\circ$$.

value of $$\cot \theta + \cosec \theta$$ = $$\cot 60^\circ + \cosec 60^\circ$$

= $$\frac{1}{\sqrt{\ 3}}+\frac{2}{\sqrt{\ 3}}$$

= $$\frac{3}{\sqrt{\ 3}}$$

= $$\sqrt{\ 3}$$