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 -}(\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=(1)/(\cos \alpha) \\ \text{hence, }(1)/(\cos\alpha)=\sqrt[]{10} \\ \cos \alpha=\frac{1}{\sqrt[]{10}} \end{gathered}`
`\begin{gathered} \text{But tan}\alpha=(\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}`