Question

question 39, 6.3.130
part 2 of 6
show that the period of f(θ) = tan θ is π.
let p(a,b) be the point on the unit - circle corresponding to θ. write the coordinates of the point q that corresponds to θ+π.
q=( - a, - b) (type an ordered pair.)
since cot θ = 1/tan θ, cot θ and tan θ have the same period. if p(x,y) is the point on the unit - circle corresponding to t, then tan t = y/x. use this definition to evaluate tan θ and tan(θ + π).
tan θ=

tan(θ + π)=

Explanation:

Step1: Recall tangent - unit - circle relationship

If $P(a,b)$ is the point on the unit circle corresponding to $\theta$, then $\tan\theta=\frac{b}{a}$ (where $a
eq0$).

Step2: Consider the point for $\theta+\pi$

The point $Q$ corresponding to $\theta + \pi$ has coordinates $(-a,-b)$. Then $\tan(\theta+\pi)=\frac{-b}{-a}$.

Step3: Simplify $\tan(\theta+\pi)$

Since $\frac{-b}{-a}=\frac{b}{a}$, we have $\tan(\theta+\pi)=\tan\theta$.

Answer:

$\tan\theta=\frac{b}{a}$
$\tan(\theta + \pi)=\frac{b}{a}$