To find the equation of the tangent, we first need to find the derivative of the curve \(y = \frac{2 - \tan x}{1 + \tan x}\) using the quotient rule.

The quotient rule states that if \(y = \frac{u}{v}\), then \(y' = \frac{v u' - u v'}{v^2}\).

Let \(u = 2 - \tan x\) and \(v = 1 + \tan x\).

Then \(u' = -\sec^2 x\) and \(v' = \sec^2 x\).

Applying the quotient rule:

\(y' = \frac{(1 + \tan x)(-\sec^2 x) - (2 - \tan x)(\sec^2 x)}{(1 + \tan x)^2}\)

\(= \frac{-(1 + \tan x)\sec^2 x - (2 - \tan x)\sec^2 x}{(1 + \tan x)^2}\)

\(= \frac{-\sec^2 x - \tan x \sec^2 x - 2\sec^2 x + \tan x \sec^2 x}{(1 + \tan x)^2}\)

\(= \frac{-3\sec^2 x}{(1 + \tan x)^2}\)

Now, substitute \(x = \frac{1}{4} \pi\) to find the gradient of the tangent:

\(\tan \left( \frac{1}{4} \pi \right) = 1\) and \(\sec^2 \left( \frac{1}{4} \pi \right) = 2\).

\(y' = \frac{-3 \times 2}{(1 + 1)^2} = \frac{-6}{4} = -\frac{3}{2}\).

The gradient of the tangent is \(-\frac{3}{2}\).

To find the y-intercept \(c\), we need the point on the curve at \(x = \frac{1}{4} \pi\):

\(y = \frac{2 - 1}{1 + 1} = \frac{1}{2}\).

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m = -\frac{3}{2}\), \(x_1 = \frac{1}{4} \pi\), and \(y_1 = \frac{1}{2}\):

\(y - \frac{1}{2} = -\frac{3}{2}(x - \frac{1}{4} \pi)\)

\(y = -\frac{3}{2}x + \frac{3}{8} \pi + \frac{1}{2}\)

Approximating \(\frac{3}{8} \pi\) to 3 significant figures gives \(1.18\).

Thus, \(y = -\frac{3}{2}x + 1.68\).