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

$$\tan \theta - \cot \theta = \cosec \theta$$

$$\tan \theta - \frac{1}{tan \theta} = \cosec \theta$$

$$\tan^2 \theta - 1 = \cosec \theta\tan \theta$$

$$\sec^2 \theta - 1 - 1 = \sec \theta$$

$$\sec^2 \theta - \sec \theta - 2 = 0$$

$$\sec^2 \theta - 2\sec \theta + \sec \theta - 2 = 0$$

$$\sec \theta(\sec \theta - 2) +1(\sec \theta - 2) = 0$$

$$ (\sec \theta + 1)(\sec \theta - 2) = 0$$

For $$0^\circ < \theta < 90^\circ$$,

$$\sec \theta$$ = 2

$$\cos \theta$$ = 1/2

$$\theta = 60\degree$$

Now,

$$\frac{2 \tan \theta - \cos \theta}{\sqrt{3} \cot \theta + \sec \theta}$$

Put the value of $$\theta$$,

= $$\frac{2 \tan 60\degree - \cos60\degree}{\sqrt{3} \cot 60\degree + \sec 60\degree}$$

= $$\frac{2\sqrt3- \frac{1}{2}}{\sqrt{3} \times \frac{1}{\sqrt3}+ 2}$$

= $$\frac{4\sqrt3- 1}{6}$$