1) cos 225° = ________ 2) tan 7π/6 = 3) tan -π/6 = ________ 4) cot -7π/6 = ________

Explanation:

Step1: Rewrite 225°

$225^{\circ}=180^{\circ} + 45^{\circ}$. According to the cosine - addition formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$, when $A = 180^{\circ}$ and $B=45^{\circ}$, $\cos(180^{\circ}+45^{\circ})=\cos180^{\circ}\cos45^{\circ}-\sin180^{\circ}\sin45^{\circ}$. Since $\cos180^{\circ}=- 1$, $\sin180^{\circ}=0$ and $\cos45^{\circ}=\sin45^{\circ}=\frac{\sqrt{2}}{2}$, we have $\cos(180^{\circ}+45^{\circ})=-1\times\frac{\sqrt{2}}{2}-0\times\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}$.

Step2: Rewrite $\frac{7\pi}{6}$

$\frac{7\pi}{6}=\pi+\frac{\pi}{6}$. According to the tangent - addition formula $\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}$, when $A=\pi$ and $B = \frac{\pi}{6}$, $\tan(\pi+\frac{\pi}{6})=\frac{\tan\pi+\tan\frac{\pi}{6}}{1-\tan\pi\tan\frac{\pi}{6}}$. Since $\tan\pi = 0$ and $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$, we get $\tan(\pi+\frac{\pi}{6})=\frac{0+\frac{\sqrt{3}}{3}}{1 - 0\times\frac{\sqrt{3}}{3}}=\frac{\sqrt{3}}{3}$.

Step3: Use the property of tangent function

The tangent function is an odd function, i.e., $\tan(-x)=-\tan x$. For $x = \frac{\pi}{6}$, $\tan(-\frac{\pi}{6})=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}$.

Step4: Rewrite $\cot(-\frac{7\pi}{6})$

First, $\cot(-\frac{7\pi}{6})=\frac{1}{\tan(-\frac{7\pi}{6})}$. And $\tan(-\frac{7\pi}{6})=-\tan\frac{7\pi}{6}=-\tan(\pi+\frac{\pi}{6})=-\frac{\sqrt{3}}{3}$. So $\cot(-\frac{7\pi}{6})=-\sqrt{3}$.

Answer:

1. $-\frac{\sqrt{2}}{2}$
2. $\frac{\sqrt{3}}{3}$
3. $-\frac{\sqrt{3}}{3}$
4. $-\sqrt{3}$