trigonometry - $sec^2theta+csc^2theta=sec^2thetacsc^2theta$

I was playing around with trigonometric functions when I stumbled across this

$$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$
Immediately I checked it to see if it was flawed so I devised a proof
$$\begin{align} \sec^2\theta+\csc^2\theta&=\sec^2\theta\csc^2\theta\\ \frac1{\cos^2\theta}+\frac1{\sin^2\theta}&=\frac1{\cos^2\theta\sin^2\theta}\\ \frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta\sin^2\theta}&=\frac1{\cos^2\theta\sin^2\theta}\\ \frac1{\cos^2\theta\sin^2\theta}&=\frac1{\cos^2\theta\sin^2\theta}\blacksquare\\ \end{align}$$
I checked over my proof many times and I couldn't find a mistake, so I assume that my claim must be true.

So my questions are:

Is there a deeper explanation into why adding the squares is the same as multiplying?

Is this just a property of these trigonometric functions or do similar relationships occur with other trigonometric functions?

And finally as an additional curiosity what does this translate into geometrically?

Answer

Any pair of trig functions (other than inverses) will give some sort of identity, by clearing denominators (if any) in the pythagorean identity.

e.g. with, $\sin$ and $\tan$ give

$$\sin^2 x + \cos^2 x = 1 \\ \sin^2 x + \left( \frac{\sin x}{\tan x} \right)^2 = 1 \\ \sin^2 x \tan^2 x + \sin^2 x = \tan^2 x \\ \tan^2 x - \sin^2 x = \sin^2 x \tan^2 x$$