# $\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$

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?

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it seems correct – dato datuashvili Aug 24 '13 at 11:16
If $a+b = 1$ and $a$ and $b$ are both nonzero, then $$a+b=1~ \Leftrightarrow ~\frac{a+b}{ab} = \frac{1}{ab} ~\Leftrightarrow \frac 1a + \frac 1b = \frac 1a \times \frac 1b.$$ – Dilip Sarwate Aug 24 '13 at 13:45

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

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

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yes if you mean that $1/cos(x)*1/sin(x)=1/(cos(x)*sin(x))$ if it is correct,then everything is ok in your calculation.adding square is multiplying what does mean? imagine

$1/3+1/4=(4+3)/12$

it is similar like this in your case

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