If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
My attempt is as follows:-
$$f(x)=\tan x\tan y$$
$$f(x)=\dfrac{\sin x\sin y}{\cos x\cos y}\tag{1}$$
Let's find out upper bound of $f(x)$, for that we need to find upper bound of $\sin x\sin y$ and lower bound of $\cos x\cos y$
$$\sin^2x+\sin^2y<1$$
$$(\sin x-\sin y)^2>=0$$
$$\sin^2x+\sin^2y-2\sin x\sin y>=0$$
$$\dfrac{\sin^2x+\sin^2y}{2}>=\sin x\sin y$$
$$\sin x\sin y<\dfrac{1}{2}\quad\forall x,y\tag{2}$$
$$(\cos x-\cos y)^2>=0$$
$$\cos^2x+\cos^2y-2\cos x\cos y>=0$$
$$1-\sin^2x+1-\sin^2y-2\cos x\cos y>=0$$
$$\dfrac{2-(\sin^2x+\sin^2y)}{2}>=\cos x\cos y$$
$$1-\dfrac{\sin^2x+\sin^2y}{2}>=\cos x\cos y$$
Now here I am stuck as I am not finding the way to calculate lower bound of $\cos x\cos y$

Using this, $$-\dfrac\pi2\le\sin^{-1}z\le\dfrac\pi2$$  for $-1\le z\le1\iff z^2\le1$
$$\sin^2y<1-\sin^2x=\cos^2x\implies\dfrac{\sin^2y}{\cos^2x}<1$$
$$\text{Similarly, }\dfrac{\sin^2x}{\cos^2y}<1$$
$$\implies\dfrac{\sin^2x\sin^2y}{\cos^2x\cos^2y}=\dfrac{\sin^2x}{\cos^2y}\cdot\dfrac{\sin^2y}{\cos^2x}<1$$
