If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

Question

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 $\cos x\cos y$

Answer 1

Using this, $$-\dfrac\pi2\le\sin^{-1}z\le\dfrac\pi2$$ for $-1\le z\le1\iff z^2\le1$

$$\sin^2y<1-\sin^2x=\cos^2x\implies\dfrac{\sin^2y}{\cos^2x}<1$$

$$\text{Similarly, }\dfrac{\sin^2x}{\cos^2y}<1$$

$$\implies\dfrac{\sin^2x\sin^2y}{\cos^2x\cos^2y}=\dfrac{\sin^2x}{\cos^2y}\cdot\dfrac{\sin^2y}{\cos^2x}<1$$