If $$12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\cosec \theta + \sec \theta}{\tan \theta + \cot \theta}$$?

$$12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3$$

$$12 \cos^2 \theta - 2(1 -  \cos^2 \theta) + 3\cos \theta = 3$$

$$14 \cos^2 \theta + 3\cos \theta = 5$$

Put the value of $$\theta = 60\degree$$,

$$14 \cos^2 60\degree + 3\cos 60\degree = 5$$

$$14 \times \frac{1}{2} + 3 \times \frac{1}{2} = 5$$

5 = 5

L.H.S. = R.H.S.

$$\frac{\cosec \theta + \sec \theta}{\tan \theta + \cot \theta}$$

= $$\frac{\cosec 60\degree + \sec 60\degree}{\tan 60\degree + \cot 60\degree}$$

= $$\frac{\frac{2}{\sqrt3} + 2}{\sqrt3 + \frac{1}{\sqrt3}}$$

= $$\frac{\frac{2 + 2\sqrt3}{\sqrt3}}{\frac{3 + 1}{\sqrt3}}$$

= $$\frac{1 + \sqrt3}{2}$$