The given differential equation is

$$x\sqrt {{x^2} - 1} dy - y\sqrt {{y^2} - 1} dx = 0$$

$$ = \int {{{dy} \over {y\sqrt {{y^2} - 1} }} = \int {{{dx} \over {x\sqrt {{x^2} - 1} }}} } $$

$$ \Rightarrow {\sec ^{ - 1}}y = {\sec ^{ - 1}}x + c$$

$$y = \sec [{\sec ^{ - 1}}x + C]$$ [$$\because$$ $$y(2) = {2 \over {\sqrt 3 }}$$]

$${2 \over {\sqrt 3 }} = \sec ({\sec ^{ - 1}}2 + C)$$

$$ = {\sec ^{ - 1}}{2 \over {\sqrt 3 }} - {\sec ^{ - 1}}2 = C$$

$$C = {\pi \over 6} - {\pi \over 3} = {{ - \pi } \over 6}$$ [$$\therefore$$ $$y = \sec \left[ {{{\sec }^{ - 1}}x - {\pi \over 6}} \right]$$]

Statement 1 is true

Also, $${1 \over y} = \cos \left[ {\cos - {\pi \over x} - {\pi \over 6}} \right]$$

$$\cos \left( {{{\cos }^{ - 1}}{1 \over x}} \right)\cos {\pi \over 6} + \sin ({\cos ^{ - 1}}{1 \over x})\sin {\pi \over 6}$$

$$ = {1 \over y} = {{\sqrt 3 } \over {2x}} + {1 \over 2}\sqrt {1 - {1 \over {{x^2}}}} $$

$$\therefore$$ Statement 2 is false.