If $$0^\circ < \theta < 90^\circ$$ and $$\cos^2 \theta = 3(\cot^2 \theta - \cos^2 \theta)$$ then the value of $$\left(\frac{1}{2} \sec \theta + \sin \theta \right)^{-1}$$ is:

From the given question,

$$\cos^2 \theta = 3(\cot^2 \theta - \cos^2 \theta)$$

$$\Rightarrow \cos^2 \theta = 3\left( \dfrac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta \right)$$

$$\Rightarrow \cos^2 \theta = 3\left( \dfrac{\cos^2 \theta- \cos^2 \theta \sin^2 \theta}{\sin^2 \theta} \right)$$

$$\Rightarrow \cos^2 \theta = 3 \cos^2 \theta\left( \dfrac{1- \sin^2 \theta}{\sin^2 \theta} \right)$$

$$\Rightarrow \cos^2 \theta = 3 \cos^2 \theta\left( \dfrac{ \cos^2 \theta}{\sin^2 \theta} \right)$$

$$\Rightarrow \cos^2 \theta \sin^2 \theta= 3 \cos^2 \theta\left( \cos^2 \theta \right)$$

$$\Rightarrow \cos^2 \theta \sin^2 \theta= 3 \cos^4 \theta$$

$$\Rightarrow 3 \cos^4 \theta -\cos^2 \theta \sin^2 \theta= 0$$

$$\Rightarrow 3 \cos^4 \theta -\cos^2 \theta(1- \cos^2 \theta)= 0$$

$$\Rightarrow 3 \cos^4 \theta -\cos^2 \theta+ \cos^4 \theta= 0$$

$$\Rightarrow 4 \cos^4 \theta -\cos^2 \theta= 0$$

$$\Rightarrow \cos^2(4 \cos^2 \theta -1)= 0$$

$$\Rightarrow 4 \cos^2 \theta -1= 0 $$ and $$\cos^2 \theta= 0$$

$$\Rightarrow 4 \cos^2 \theta= 1 $$ and $$\cos^2 \theta= 0$$

$$\Rightarrow \cos^2 \theta= \dfrac{1}{4} $$ and $$\cos \theta=\cos( \dfrac{ \pi}{2})= 0$$

$$\Rightarrow \cos \theta= \dfrac{1}{2}=\cos{\dfrac{\pi}{3}} $$ and $$\cos \theta=\cos( \dfrac{ \pi}{2})= 0$$

Hence, $$\theta=\dfrac{\pi}{3}$$ And $$\theta=\dfrac{\pi}{2}$$

As per the question it is given that $$ 0<\theta<\dfrac{\pi}{2}$$

Now, substituting the values of $$\theta=\dfrac{\pi}{3}$$

$$\Rightarrow \left(\frac{1}{2} \sec (\dfrac{\pi}{3}) + \sin (\dfrac{\pi}{3}) \right)^{-1}$$

$$\Rightarrow \left(\frac{1}{2} 2+ \dfrac{\sqrt 3}{2} \right)^{-1}$$

$$\Rightarrow \left(1+ \dfrac{\sqrt 3}{2} \right)^{-1}$$

$$\Rightarrow \left( \dfrac{2+\sqrt 3}{2} \right)^{-1}$$

$$\Rightarrow \left( \dfrac{2}{2+\sqrt 3} \right)$$

Now, taking the conjugate of the denominator,

$$\Rightarrow \left( \dfrac{2(2-\sqrt 3)}{(2+\sqrt 3)(2-\sqrt 3)}\right)= \dfrac{2(2-\sqrt 3)}{(4- 3)}= 2(2-\sqrt 3)$$