[tex]y = x + \pi[/tex]
Step-by-step explanation:
We need to find the slope of the line tangent to y at [tex]x=-\pi[/tex] and we can do that by calculating the derivative of y at that point. We know that
[tex]y'(x) = \dfrac{d}{dx}(\tan{x}) = \sec^2{x}[/tex]
and
[tex]y'(-\pi) = \sec^2(-\pi) = 1[/tex]
To find the equation of the tangent line, we know that the line passes through the point [tex](-\pi, 0)[/tex] so we can write the slope-intercept form of the line's equation as
[tex]0 = (-\pi) + b \Rightarrow b = \pi[/tex]
Therefore, the equation of the line tangent to the point [tex]x = -\pi[/tex] is
[tex]y = x + \pi[/tex]
The graph of the two functions is shown above.
