If $$\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan \theta + \sec \theta)^{-1}$$ is:

$$\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}$$
Put the $$\theta = 60\degree$$,
$$\frac{\sin 60\degree}{1 + \cos 60\degree} + \frac{1 + \cos 60\degree}{\sin 60\degree} = \frac{4}{\sqrt{3}}$$
$$\frac{\frac{\sqrt{3}}{2}}{1 +\frac{1}{2}} + \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}}$$
$$\frac{\frac{3}{4} + \frac{9}{4}}{\frac{3}{2} \times \frac{3}{4}} = \frac{4}{\sqrt{3}}$$
$$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$$
Now,
$$(\tan \theta + \sec \theta)^{-1}$$
Put the $$\theta = 60\degree$$,
= $$(\tan 60\degree + \sec 60\degree)^{-1}$$
= $$(\sqrt{3} + 2)^{-1}$$
= $$2 - \sqrt{3}$$