The equation of a tangent to the hyperbola, $$4x^2 - 5y^2 = 20$$, parallel to the line $$x - y = 2$$, is:

We are asked to write the equation of a tangent to the hyperbola $$4x^2-5y^2=20$$ that is parallel to the straight line $$x-y=2$$.

First we note that a line written as $$x-y=2$$ can be rearranged as $$y=x-2.$$ From this form we immediately read its slope; the coefficient of $$x$$ is $$1$$, so the slope of every line parallel to it is also $$m=1$$.

Now we place the hyperbola in its standard form. Dividing every term of $$4x^2-5y^2=20$$ by $$20$$, we get $$\frac{x^2}{5}-\frac{y^2}{4}=1.$$ Comparing with the general standard form $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ we identify $$a^2=5 \quad\text{and}\quad b^2=4.$$ Hence $$a=\sqrt5,\qquad b=2.$

The slope form of the tangent to a hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$ is a known result; it is

$$y = mx \;\pm\; \sqrt{a^2m^2-b^2}.$$

We explicitly cite this formula so that each subsequent substitution is perfectly clear.

We already found the desired slope $$m=1$$. Substituting $$m=1$$, $$a^2=5$$, and $$b^2=4$$ into the square-root term, we proceed one algebraic step at a time:

$$a^2m^2-b^2 =5\,(1)^2-4 =5-4 =1.$$

The square root of $$1$$ is simply $$1$$, so the tangent lines become

$$y = 1\cdot x \;\pm\; 1,$$

or written more transparently,

$$y = x + 1 \quad\text{and}\quad y = x - 1.$$

To match the style of the answer choices, we transfer each line to the form $$Ax + By + C = 0$$ by moving every term to one side:

For $$y = x + 1$$ we subtract $$y$$ and add $$1$$ to the left:

$$x - y + 1 = 0.$$

For $$y = x - 1$$ we again subtract $$y$$ and now subtract $$1$$ on the left:

$$x - y - 1 = 0.$$

Among the four answer choices given in the problem statement we find only one of these two candidates, namely

$$x - y + 1 = 0.$$

Therefore the required tangent, parallel to the line $$x-y=2$$, is $$x - y + 1 = 0$$.

Hence, the correct answer is Option C.