If $$x \cos A - y \sin A = 1$$ and $$x \sin A + y \cos A = 4$$, then the value of $$17x^{2} + 17y^{2}$$ is:

$$x \cos A - y \sin A=1$$

Take the square of both sides,

$$(x \cos A - y \sin A)^2=1$$

$$(x \cos A)^2 - 2xy \sin A \cos A + (y \sin A)^2 = 1$$ ---(1)

($$\because (a + b)^2 = a^2 + b^2 + 2ab$$)

$$x \sin A + y \cos A = 4$$

Take the square of both sides,

$$(x \sin A + y \cos A)^2=16$$

$$(x \sin A)^2 + 2xy \sin A \cos A + (y \cos A)^2 =16$$ ---(2)

Addition of eq(1) and eq(2),

$$(x \cos A)^2 + (x \sin A)^2 + (y \sin A)^2 + (y \cos A)^2 = 17$$

$$x^2 + y^2 = 17$$

$$17x^{2}+17y^{2} = 17 \times 17$$

$$17x^{2}+17y^{2} = 289$$