<s>[INST] Simplify $\frac{4\sin\alpha-5\cos\alpha}{3\cos\alpha-\sin\alpha}$ Simplify $\dfrac{4\sin\alpha-5\cos\alpha}{3\cos\alpha-\sin\alpha}.$
I think we should get $\cot\alpha$ which is $\dfrac{\cos^2\alpha}{\sin^2\alpha}.$ So I tried to write $\sin\alpha$ as $\sqrt{1-\cos^2\alpha}$ but it didn't work.
 [/INST] Using Weierstrass relations for double angles. Let
$$ \alpha = 2 \beta , t = \tan \beta,\; \sin \alpha= \dfrac{2t}{1+t^2}=S,\; \cos \alpha= \dfrac{1-t^2}{1+t^2}=C;\;$$
When we simplify
$$ \dfrac{4C-5S}{3S-C}$$
we get in terms of tan half angle $t$
$$-\dfrac{5t^2+8t-5}{3t^2+2t-3}$$
but not
$$ \dfrac{1-t^2}{2t}= \cot \alpha $$
as you are expecting.
</s>