Question

Hi i don’t know how to answer #4 and #5. I’m in college calculus 1.

Answer 1

$\begin{gathered} \text{Given tan}\alpha=-3\text{ with -}\frac{\pi}{2}<\alpha<0 \\ \end{gathered}$$\tan ^2\theta+1=\sec ^2\theta_{}$

Hence,

$\begin{gathered} \tan ^2\alpha+1=\sec ^2\alpha \\ \text{Substitute tan}\alpha=-3\text{ into the eqaution above},\text{ we have } \\ (-3)^2+1=\sec ^2\alpha \\ 9+1=\sec ^2\alpha \\ 10=\sec ^2\alpha \\ \sec ^2\alpha=10 \\ \sec \alpha=\sqrt[]{10} \\ \text{but sec}\alpha=\frac{1}{\cos \alpha} \\ \text{hence, }\frac{1}{\cos\alpha}=\sqrt[]{10} \\ \cos \alpha=\frac{1}{\sqrt[]{10}} \end{gathered}$$\begin{gathered} \text{But tan}\alpha=\frac{\sin \alpha}{\cos \alpha} \\ \sin \alpha=\tan \alpha\cos \alpha \\ \sin \alpha=-3\text{ x }\frac{1}{\sqrt[]{10}} \\ \sin \alpha=-\frac{3}{\sqrt[]{10}} \\ \\ Hence,\text{ }\sin \alpha=-\frac{3}{\sqrt[]{10}} \end{gathered}$