Verified answer

y - 1 = f '(pi/8) (x - (pi/8))

The derivative of tanx is 1/cos^2x.

By the chain rule, the derivative (and thus the slope of the tangent) is (2)(1/cos^2x) = 2/cos^2x

Now at (pi/8, 1) using slope and y0intercept form of a line,

y=mx+b

Differentiate the function:

f'(x) = sec^2(2x) * 2

f'(x) = 2sec^2(2x)

Sub in the given coordinates to find the slope of the tangent line:

f'(x) = 2/cos^2(π/4)

f'(x) = 4

Now that you have the slope of the tangent line, you can find its equation:

y - 1 = 4(x - π/8)

y - 1 = 4x - π/2

8x - 2y = π - 2