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9-12.

Multiple Choice: The equation of the line tangent to $y =\operatorname{tan}^{−1}x$ at the point where $x = \sqrt { 3 }$ is:

$y = \frac { 1 } { 4 } x - \frac { \sqrt { 3 } } { 4 } + \frac { \pi } { 3 }$

$y = \frac { 1 } { 4 } x - \frac { \sqrt { 3 } } { 4 } + \frac { \pi } { 6 }$

$y = - \frac { 1 } { 2 } x + \frac { \sqrt { 3 } } { 2 } + \frac { \pi } { 3 }$

$y = - \frac { 1 } { 2 } x + \frac { \sqrt { 3 } } { 2 } + \frac { \pi } { 6 }$

Hint:

The point of tangency is:
Use the slope and the point of tangency to write the equation of the line.

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