Find the equation of the tangent line to the curve y=5sec(x)-10cos(x) at the point (pi/3, 5) written in the form y=mx+b, where m=? and b=?

Y = 5secx - 10cosx 
y' = 5secxtanx + 10sinx 

m at (3, 5) = 5sec(π/3)tan(π/3) + 10sin(π/3) = 5(2)(√3) + 10(√3/2) 
m = 6√3 + 5√3 = 11√3 

y - 5 = 11√3(x - π/3) 
y = 11√3 x - 33π√3 + 5

The equation of a line is normally of the form:
 y = mx + b ; where m = slope and b = y integer

The derivative of a function is the slope at any given point so:
m = f'(x); and the point itself is used to calibrate the rest

Tangent line = f'(x0) (x -x0) + y0 ;
where x0 and y0 are the components for the point being used

So,
m = f'(pi/3)
    = f'(1.04)  

b = -f'(pi/3)*(pi/3) + 5
   =-f'(1.0816) +5

Dis a correct answer