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 $\cos\alpha=\pm\dfrac{12}{13}$.
Best Answer
$$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(1-2\sin^2\frac{\alpha}2\right)}=\dfrac{4-10\sin\frac{\alpha}2\cos\frac{\alpha}2}{5-6\sin^2\frac{\alpha}2}$$Now divide numerator and denominator by $\sin^2\frac{\alpha}2$, $$\dfrac{\dfrac{4}{\sin^2\frac{\alpha}2}-10\cot\frac{\alpha}2}{\dfrac5{\sin^2\frac{\alpha}2}-6}=\dfrac{4(1+\cot^2\frac{\alpha}2)-10\cot\frac{\alpha}2}{5(1+\cot^2\frac{\alpha}2)-6}=\frac{112}{41}$$