$\text { What is } P _ { 1 } \text {, a first-degree polynomial function whose value and slope agree with the }$ value and slope of $f ( x ) = 10 \tan ( x )$ at $x = \frac { \pi } { 4 }$ ?
A) $P _ { 1 }$ is the tangent line to the curve $f ( x ) = 10 \tan ( x )$ at the point $\left( 10 , \frac { \pi } { 4 } \right)$.
B) $P _ { 1 }$ is the tangent line to the curve $f ( x ) = 10 \tan ( x )$ at the point $\left( \frac { \pi } { 4 } , 10 \right)$.
C) $P _ { 1 }$ is the tangent line to the curve $f ( x ) = 10 \tan ( x )$ at the point $\left( 13 , \frac { \pi } { 4 } \right)$.
D) $P _ { 1 }$ is the tangent line to the curve $f ( x ) = 10 \tan ( x )$ at the point $\left( 10 , - \frac { \pi } { 4 } \right)$.
E) $P _ { 1 }$ is the tangent line to the curve $f ( x ) = 10 \tan ( x )$ at the point $\left( \frac { \pi } { 4 } , - 10 \right)$.

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to $f(x)=10\tan(x)$ at $x=\frac{\pi}{4}$, we first need to find the value of $f(\frac{\pi}{4})$ and $f'(\frac{\pi}{4})$.
We have $f(\frac{\pi}{4})=10\tan(\frac{\pi}{4})=10\cdot 1=10$.
And $f'(x)=10\sec^2(x)$, so $f'(\frac{\pi}{4})=10\sec^2(\frac{\pi}{4})=10\cdot 2=20$.
Now we can use the point-slope form of the equation of a line:
$$y-y_1=m(x-x_1)$$
where $(x_1,y_1)$ is the point of tangency and $m$ is the slope of the tangent line.
Plugging in the values we found, we get:
$$y-10=20(x-\frac{\pi}{4})$$
Simplifying and rearranging, we get:
$$y=20x-2\pi+10$$
Therefore, the correct choice is B, as it gives the coordinates of the point of tangency.

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