What is the equation of the line tangent to f(x)=4sec⁡x−8cos⁡x at x=3?

Equation of line tangent can be formed by ${\displaystyle y_1-y=m(x_1-x)}$ where m is the slope and ${\displaystyle x_1}$ and ${\displaystyle y_1}$ represent the x-coordinate and y-coordinate of the point of intersection of f(x) and tangent.
We know that slope of line tangent to y=f(x) is ${\displaystyle f^{\prime }\left(x\right)=\frac{dy}{dx}}$.
So, to do this question, we have to find f'(x) first.
${\displaystyle f\left(x\right)=4\sec x-8\cos x}$
${\displaystyle f^{\prime }\left(x\right)=4\sec x\tan x-8(-\sin x)}$
${\displaystyle =4\sec x\tan x+8\sin x}$
When x=3, the slope of line tangent (m)=f'(3)
${\displaystyle =4\sec \left(3\right)\tan \left(3\right)+8\sin \left(3\right)}$
and the y-coordinate of this point is f(3)
${\displaystyle =4\sec \left(3\right)-8\cos \left(3\right)}$
We get all the information we need and we can plug them into the equation.
Equation of line tangent at x=3:
${\displaystyle 4\sec \left(3\right)-8\cos \left(3\right)-y=[4\sec \left(3\right)\tan \left(3\right)+8\sin \left(3\right)](3-x)}$