find the equation of the tangent line at f(x)=tanx - 2 at point (pie/8,3)
f = tanx - 2
f' = sec2x
f'(π/8) = 1.1715
Unfortunately, f(π/8) is not 3, but rather √2 - 4 = -2.5857
However, if we assume that (π/8,3) is our point of tangency, then the line through (π/8,3) with slope 1.1715 is
y-3 = 1.1715(x-π/8)
Adjust to fit facts as you see fit.