Find value of $\cot(\theta-\alpha)$ Let $\theta$ and $\alpha$ be the solutions to $$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=\dfrac{3\cos2x+6}{10}$$
find the value of $\cot(\theta-\alpha)$, where $\dfrac{\pi}{2}<\theta<\alpha<\dfrac{3\pi}{2}$
My attempt :
Let $A=\dfrac{x}{2}$, $$\dfrac{\sin^2A}{1-\cot A}+\dfrac{\cos^2A}{1-\tan A}=\dfrac{3\cos 4A+6}{10}$$
$$\dfrac{\sin A\cos A-\sin^4A-\cos^4A}{2\sin A\cos A-\cos^2A-\sin^2A}=\dfrac{12\cos^2A-3}{10}$$
$$\dfrac{\sin A\cos A+\cos^2A-\sin^2A}{2\sin A\cos A-1}=\dfrac{12\cos^3A-3}{10}$$
Are there any better ways for approaching this problem? or I've come the right way.

Hint:
For $\sin \dfrac x2=s,\cos\dfrac x2=c$ and $c\ne s$
$$\dfrac{s^2}{1-\dfrac cs}+\dfrac{c^2}{1-\dfrac sc}=\dfrac{s^3-c^3}{s-c}=\cdots=1+\dfrac{\sin x}2$$
Now use $\cos2x=1-2\sin^2x$ to form a Quadratic Equation in $\sin x$
