On the following unit circle, $ heta$ is in radians and $ an ( heta)=\frac{-0.99}{0.1}=-9.9$.

Without a calculator, evaluate the following expressions to the nearest hundredth or as an exact ratio.
\begin{tabular}{ll}
Expression & Value \
\hline $ an (\pi+ heta)$ & $\square$ \
$ an (2 \pi+ heta)$ & $\square$
\end{tabular}

Q: On the following unit circle, $\theta$ is in radians and $\tan (\theta)=\frac{-0.99}{0.1}=-9.9$. Without a calculator, evaluate the following expressions to the nearest hundredth or as an exact ratio. \begin{tabular}{ll} Expression & Value \\ \hline $\tan (\pi+\theta)$ & $\square$ \\ $\tan (2 \pi+\theta)$ & $\square$ \end{tabular}

∻Solution by Steps∻ ‖ step 1 ⋮ To find $tan(\pi + \theta)$, we use the periodic property of the tangent function, which states that $tan(\pi + \theta) = tan(\theta)$. ‖ ‖ step 2 ⋮ Since $tan(\theta) = -9.9$, we have $tan(\pi + \theta) = -9.9$. ‖ ∻1 Answer∻ ⚹ $tan(\pi + \theta) = -9.9$ ⚹ ‖ step 1 ⋮ To find $tan(2\pi + \theta)$, we use the periodic property of the tangent function, which states that $tan(2\pi + \theta) = tan(\theta)$. ‖ ‖ step 2 ⋮ Since $tan(\theta) = -9.9$, we have $tan(2\pi + \theta) = -9.9$. ‖ ∻2 Answer∻ ⚹ $tan(2\pi + \theta) = -9.9$ ⚹ ∻Key Concept∻ ⚹Periodicity of the Tangent Function⚹ ∻Explanation∻ ⚹The tangent function is periodic with a period of $\pi$, which means $tan(\theta) = tan(\theta + n\pi)$ for any integer $n$. Therefore, $tan(\pi + \theta)$ and $tan(2\pi + \theta)$ both simplify to $tan(\theta)$.⚹◊What is the value of $tan(\pi + \theta)$ when evaluated without a calculator?⍭ Generate me a similar question◊

Question

On the following unit circle, $	heta$ is in radians and $	an (	heta)=\frac{-0.99}{0.1}=-9.9$.

Without a calculator, evaluate the following expressions to the nearest hundredth or as an exact ratio.
\begin{tabular}{ll} 
Expression & Value \
\hline $	an (\pi+	heta)$ & $\square$ \
$	an (2 \pi+	heta)$ & $\square$
\end{tabular}

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $tan(\pi + heta)$, we use the periodic property of the tangent function, which states that $tan(\pi + heta) = tan( heta)$. ‖ ‖ step 2 ⋮ Since $tan( heta) = -9.9$, we have $tan(\pi + heta) = -9.9$. ‖ ∻1 Answer∻ ⚹ $tan(\pi + heta) = -9.9$ ⚹ ‖ step 1 ⋮ To find $tan(2\pi + heta)$, we use the periodic property of the tangent function, which states that $tan(2\pi + heta) = tan( heta)$. ‖ ‖ step 2 ⋮ Since $tan( heta) = -9.9$, we have $tan(2\pi + heta) = -9.9$. ‖ ∻2 Answer∻ ⚹ $tan(2\pi + heta) = -9.9$ ⚹ ∻Key Concept∻ ⚹Periodicity of the Tangent Function⚹ ∻Explanation∻ ⚹The tangent function is periodic with a period of $\pi$, which means $tan( heta) = tan( heta + n\pi)$ for any integer $n$. Therefore, $tan(\pi + heta)$ and $tan(2\pi + heta)$ both simplify to $tan( heta)$.⚹◊What is the value of $tan(\pi + heta)$ when evaluated without a calculator?⍭ Generate me a similar question◊

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