<s>[INST] Calculate $\frac{4-5\sin\alpha}{2+3\cos\alpha}$ 
Calculate $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}$$ if $\cot\dfrac{\alpha}{2}=-\dfrac32$.

My first approach was to somehow write the given expression only in terms of the given $\cot\frac{\alpha}{2}$ and just put in the value $\left(-\dfrac{3}{2}\right)$. Now I don't think that's possible because we have constants (4 and 2). My try, though: $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}\right)}=\dfrac{4-10\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\cos^2\frac{\alpha}{2}-3\sin^2\frac{\alpha}{2}}$$ My second idea was to find the value of the trig functions of $\alpha$. I don't know if this is the most straight-forward approach, but $$\cot\alpha=\dfrac{\cot^2\frac{\alpha}{2}-1}{2\cot\frac{\alpha}{2}}=\dfrac{\frac94-1}{-2.\frac32}=-\dfrac{5}{12}.$$ Am I now supposed just to find the values of $\sin\alpha$ and $\cos\alpha$? Nothing more elegant? We would have $\dfrac{\cos\alpha}{\sin\alpha}=-\dfrac{5}{12}\Rightarrow\cos\alpha=-\dfrac{5}{12}\sin\alpha$ and putting into $\sin^2\alpha+\cos^2\alpha=1$ we'd get $\cos\alpha=\pm\dfrac{12}{13}$.
 [/INST] We have $$\cot{\frac{\alpha}{2}}=\frac{\sin \alpha}{1-\cos \alpha}=-\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=-\frac{3}{2} \implies \sin \alpha=1.5\cos \alpha -1.5;$$ $$3\sqrt{1-\cos \alpha}=2\sqrt{1+\cos \alpha}$$ 
Thus, $9(1-\cos \alpha)=4(1+\cos \alpha) \implies \cos \alpha= \frac{5}{13}$
$$\frac{4-5\sin \alpha}{2+3\cos \alpha}=\frac{4-7.5\cos \alpha+7.5}{2+3\cos \alpha}=\frac{112}{41}$$
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