You know that the f'(x) is the slope of the tangent line to anywhere on the curve of the f(x).

So if you can find the particular point on the curve where the tangent line touches the curve, then just find that point. Then solve for the f'(x) at that point. Then use that point and the slope of the tangent line there to find the equation of the said tangent line by using the point-slope form of the equation of a line:

(y -y1) = m(x -x1) --------(i)

The problem says the tangent is at -pi/4.

That could mean it is at x = -pi/4.

(It could also mean it is at y = -pi/4, or f(x) = -pi/4....though I doubt that.)

So you have x1 = -pi/4

What would be the y1?

It is f(-pi/4).

It is y1 = tan(-pi/4) -cot(-pi/4) = 0

So, point(-pi/4,0).

The slope of the tangent line there is

m = f'(-pi/4) = sec^2(-pi/4) +csc^2(-pi/4) = 4

Hence,

(y -0) = 4(x -(-pi/4))

y = 4(x +pi/4)

y = 4x +pi ---------the tangent line.