If $0<x<90$ and $0<y<90$, why does $\tan\left(x\right)\tan\left(y\right)< 1 \implies x + y < 90$?
I know that $\tan{x}\tan{y} = \frac{\sin{x}\sin{y}}{\cos{x}\cos{y}}$, but I don't see if $\frac{\sin{x}\sin{y}}{\cos{x}\cos{y}}<1\implies x + y < 90$ is any easier to make sense of.
From the constraints $0 < x < 90°$ and $0 < y < 90°$, we know that $0 < x + y < 180°$, and $\sin x$, $\cos x$, $\sin y$, and $\cos y$ are all positive.
$$\tan(x) \tan(y) < 1$$ $$\frac{\sin(x)\sin(y)}{\cos(x)\cos(y)} < 1$$ $$\sin(x)\sin(y) < \cos(x)\cos(y)$$ $$0 < \cos(x)\cos(y) - \sin(x)\sin(y)$$ $$0 < \cos(x + y)$$ $$\cos(x + y) > 0$$ $$x + y < 90°$$
$$tan(x+y)=\frac{tanx + tany}{1-tanx\cdot tany}$$ Now due to given domains of $x, y$ we have $tanx\gt 0, tany\gt 0$. Also given is $tanx \cdot tany \lt 1$, so $tan (x+y)\gt 0$.
Now,
$tan\theta$ is positive in the first and third quadrants, but since $0^{\circ}<x,y<90^{\circ}$, Thus $ 0^{\circ}<x+y<180^{\circ}$. This limits the range of x+y to $(0^{\circ},90^{\circ})$.
Let $x \in (0,90)$ and $z \in (0,90)$. If $\tan x \tan z = 1$, then $\tan x = \cot z$. Consider now a right angle triangle with angles $90$ degrees, $x$, and another angle. By definition, this another angle will be $z$ so that $x+z=90$.
Now, fix arbitrary $x \in (0,90)$. If $\tan x \tan y < 1$ for some $y \in (0,90)$, then $\tan x < \cot y$. Since $\cot$ is strictly decreasing in the interval $(0,90)$, we have $y < z=90-x$ by looking at the right angle triangle described above. Therefore, $x+y<90$.
Use angle sum identities to establish that
$$\tan(x)\tan(y) = \frac{1-\cos(x+y)/\cos(x-y)}{1+\cos(x+y)/\cos(x-y)}.$$
Because $\cos(\cdot)$ is an even function, assume WLOG that $x>y$.
Then, $\cos(x-y)>0$. For the fraction to be less than $1$, $\cos(x+y)>0$ which implies that
$0<x+y<90$.
