Statement-1: Two tangents drawn from any point on he hyperbola $x^2-y^2=a^2-b^2\text{ to the ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}$ =1 make complementary angles with the axis of the ellipse.

Statement-2: If two lines make complementary angles with the axis of x then the product of their slopes is 1.

If the angle in statement-2 are $\alpha \text{ and }\beta$ such that $\alpha +\beta =\pi /2$ then the product of the slopes is tan $\alpha \tan \beta$ = 1, and the statement-2 is true.

In statement-1, any tangent to the ellipse is

$y=mx+\sqrt{a^2{m}^2+b^2}$ which passes through

$\begin{array}{l}\left(\sqrt{a^2-b^2}\sec \theta ,\sqrt{a^2-b^2}\tan \theta \right)\\ \Rightarrow \left(a^2-b^2\right)(\tan \theta -m\sec \theta {)}^2=a^2{m}^2+b^2\end{array}$

Product of the slopes 

$=\frac{\left(a^2-b^2\right){\tan }^2\theta -b^2}{\left(a^2-b^2\right){\sec }^2\theta -a^2}=\frac{a^2{\sin }^2\theta -b^2}{a^2{\sin }^2\theta -b^2}=1$

So by statement-2, statement-1 is also true.