JEE Math Practice Question

Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. Then the area of the region bounded by the curves x = 0, $$x = {1 \over {\sqrt 2 }}$$ and y = y(x) in the upper half plane is :

1. A.$${1 \over 8}(\pi - 1)$$
2. B.$${1 \over {12}}(\pi - 3)$$
3. C.$${1 \over 4}(\pi - 2)$$
4. D.$${1 \over 6}(\pi - 1)$$