Exact value of $\frac{\arccos(1-2\tan^2\alpha)}{2\arcsin(\tan\alpha)}$

Question

Let $\alpha\in\left(0,\dfrac\pi2\right)$. What is the exact value of $$\dfrac{\arccos(1-2\tan^2\alpha)}{2\arcsin(\tan\alpha)}$$ Firstly, I tried to simplify $1-2\tan^2\alpha$ and got $$\dfrac{3\cos^2\alpha-2}{\cos^2\alpha}$$ What is next step? Is there a formula to simplify $\arccos\dfrac{a}{b}$ for some $a,b\in\mathbb{R}$?

Answer 1

Let $\theta = \arcsin (\tan \alpha) \to \sin \theta = \tan \alpha \to 1 - 2\tan^2\alpha = 1 - 2\sin^2\theta = \cos 2\theta \to \arccos \left(1-2\tan^2\alpha\right) = \arccos (\cos 2\theta) = 2\theta \to L = \dfrac{2\theta}{2\theta} = 1$