y = mx + b for a tangent line

Question

can someone help me find the equation of the tangent line for the equation: y = 5 sec x - 10 cos x at the point, (pi/3, 5)

the tangent line equation is: (15sqrt3)x - (5sqrt3) pi + 5 , however the tangent line has to be in the form of y = mx + b

can someone help me put the equation into this form?

Answer 1

f'(x) = 5 sec(x)tanx(x) + 10 sin(x)

Evaluating f'(x) at x = pi/3 yields

f'(pi/3) = 5*2*sqrt(3) + 5sqrt(3) = 15sqrt(3)

y = mx + b. Your point is (pi/3, 5) and your m is 15sqrt(3)

5 = 15sqrt(3)*(pi/3) + b

b = 5 - 15sqrt(3)*(pi/3)

Therefore y = 15sqrt(3)x + [5 - 15sqrt(3)*(pi/3)]