Question

Find tan 0 for each value of 0Try these C-D

Answer 1

We have to find the value of the tangent function for these angles:

a. θ = 300°

$\begin{gathered} \tan (300\degree)=\tan (300\cdot\frac{\pi}{180})=\tan (\frac{10}{6}\pi)=\tan (\frac{4}{6}\pi+\frac{6}{6}\pi)=\tan (\frac{2\pi}{3}) \\ \tan (\frac{2\pi}{3})=\frac{\sin(\frac{2\pi}{3})}{\cos(\frac{2\pi}{3})}=\frac{\frac{\sqrt[]{3}}{2}}{-\frac{1}{2}}=-\sqrt[]{3} \end{gathered}$

b. θ = 450°

$\begin{gathered} \tan (450\degree)=\tan (450\cdot\frac{\pi}{180})=\tan (\frac{5}{2}\pi)=\tan (\frac{1}{2}\pi+2\pi)=\tan (\frac{\pi}{2}) \\ \tan (\frac{\pi}{2})=\frac{\sin (\frac{\pi}{2})}{\cos (\frac{\pi}{2})}=\frac{1}{0}=\text{undefined} \end{gathered}$

c. θ = 2π/3 = 300°

$\tan (\frac{2\pi}{3})=-\sqrt[]{3}$

d. θ = 11π/4

$\begin{gathered} \tan (\frac{11\pi}{4})=\tan (\frac{3\pi}{4}+2\pi)=\tan (\frac{3\pi}{4}) \\ \tan (\frac{3\pi}{4})=\frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})}=\frac{\frac{\sqrt[]{2}}{2}}{-\frac{\sqrt[]{2}}{2}}=-1 \end{gathered}$

Answer:

a. -√3

b. Undefined

c. -√3

d. -1