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=?

To find the equation of the tangent line to the curve at a specific point, you need to determine the slope of the tangent line, which is given by the derivative of the function at that point, and then use the point-slope form to construct the equation of the line.

First, we need to find the derivative of the function y with respect to x.

Given the function: y = 5sec(x) - 10cos(x)

The derivative y' (also written as dy/dx), using the derivatives of sec(x) and cos(x), is: y' = 5sec(x)tan(x) + 10sin(x)

Next, we evaluate the derivative at the point (π/3, 5). Recall that sec(x) is 1/cos(x) and tan(x) is sin(x)/cos(x). At x = π/3, we have: cos(π/3) = 1/2, so sec(π/3) = 2 sin(π/3) = √3/2, so tan(π/3) = sin(π/3)/cos(π/3) = (√3/2)/(1/2) = √3

Now plug these values into the derivative: y' = 5(2)(√3) + 10(√3/2) y' = 10√3 + 5√3 y' = 15√3

The slope of the tangent line at (π/3, 5) is 15√3.

Now use the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line (given as (π/3, 5) in the question).

Since we already have the slope m = 15√3 and the point (π/3, 5), plug these into the equation: y - 5 = 15√3(x - π/3)

Now solve for y to put the equation in slope-intercept form y = mx + b: y = 15√3(x - π/3) + 5 y = 15√3x - 15√3(π/3) + 5 y = 15√3x - 5π√3 + 5

So the equation of the tangent line is: y = 15√3x - 5π√3 + 5

To identify m and b explicitly: m = 15√3 b = -5π√3 + 5