Question

10.已知$\tan\alpha=-3$,则$\dfrac{\sin\alpha+2\cos\alpha}{\sin\alpha-\cos\alpha}=()$
a.$\dfrac{5}{2}$ b.$\dfrac{1}{4}$
c.$\dfrac{5}{4}$ d.$\dfrac{7}{2}$

Explanation:

Step1: Divide numerator and denominator by $\cos\alpha$

Since $\cos\alpha
eq0$ (otherwise $\tan\alpha$ would be undefined), we divide both the numerator and the denominator of the fraction $\frac{\sin\alpha + 2\cos\alpha}{\sin\alpha-\cos\alpha}$ by $\cos\alpha$. We know that $\frac{\sin\alpha}{\cos\alpha}=\tan\alpha$, so we get:
$\frac{\frac{\sin\alpha}{\cos\alpha}+2\frac{\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha}{\cos\alpha}-\frac{\cos\alpha}{\cos\alpha}}=\frac{\tan\alpha + 2}{\tan\alpha-1}$

Step2: Substitute $\tan\alpha = - 3$

We are given that $\tan\alpha=-3$, substitute this value into the expression $\frac{\tan\alpha + 2}{\tan\alpha-1}$:
$\frac{-3 + 2}{-3-1}=\frac{-1}{-4}=\frac{1}{4}$? Wait, no, wait, let's recalculate. Wait, numerator: $\tan\alpha + 2=-3 + 2=-1$? No, wait, no, wait the original numerator after division is $\tan\alpha+2$, denominator is $\tan\alpha - 1$. Wait, $\tan\alpha=-3$, so numerator: $-3 + 2=-1$? No, wait, no, I made a mistake. Wait, the original fraction is $\frac{\sin\alpha+2\cos\alpha}{\sin\alpha - \cos\alpha}$, when we divide numerator and denominator by $\cos\alpha$, we have $\frac{\frac{\sin\alpha}{\cos\alpha}+\frac{2\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha}{\cos\alpha}-\frac{\cos\alpha}{\cos\alpha}}=\frac{\tan\alpha + 2}{\tan\alpha-1}$. Now substitute $\tan\alpha=-3$: $\frac{-3 + 2}{-3-1}=\frac{-1}{-4}=\frac{1}{4}$? But that's option B? Wait, no, wait, maybe I messed up the signs. Wait, $\tan\alpha=-3$, so $\frac{\tan\alpha + 2}{\tan\alpha - 1}=\frac{-3 + 2}{-3 - 1}=\frac{-1}{-4}=\frac{1}{4}$. Wait, but let's check again. Wait, the numerator is $\sin\alpha+2\cos\alpha$, denominator $\sin\alpha - \cos\alpha$. If $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-3$, so $\sin\alpha=-3\cos\alpha$. Substitute into the fraction: $\frac{-3\cos\alpha+2\cos\alpha}{-3\cos\alpha-\cos\alpha}=\frac{(-3 + 2)\cos\alpha}{(-3 - 1)\cos\alpha}=\frac{(-1)\cos\alpha}{(-4)\cos\alpha}=\frac{-1}{-4}=\frac{1}{4}$. So the answer is B.

Answer:

B. $\frac{1}{4}$