<s>[INST] $\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$ 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?
 [/INST] 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
$$ \sin^2 x + \cos^2 x = 1 
\\ \sin^2 x + \left( \frac{\sin x}{\tan x} \right)^2 = 1
\\ \sin^2 x \tan^2 x + \sin^2 x = \tan^2 x
\\ \tan^2 x - \sin^2 x = \sin^2 x \tan^2 x
$$
</s>