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One has that the sum of the squares of the functions sine, cosine and tangent satisfy

$$ \sin^2(x) + \cos^2(x) + \tan^2(x) = \frac{d}{dx} \tan (x)$$

Does this have any nice interpretation or meaning, or is it simply just a complicated way of writing $1 + \tan^2(x)$ without given any insight?

Michael Hardy
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    I would say the second statement of yours is true. – SchrodingersCat Dec 24 '15 at 19:01
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    Well, you could see this as meaning that the rate of change of the tangent function is equal to the square of the diagonal of the parallelepiped with sides $\sin, \cos, \tan$. Why (geometrically) that holds is not clear to me, however. – Aloizio Macedo Dec 25 '15 at 00:46

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It's just an equality as both the sides are equal to $\sec^2 x$ $$LHS=\sin^2 x+\cos^2 x+\tan^2 x=(\sin^2x+\cos^2x)+\tan^2x=1+\tan^2x=\sec^2x$$ $$RHS=\frac{d}{dx}(\tan x)=\sec^2x$$ so briefly, $$LHS=RHS=\sec^2 x$$ in other words, it just a way rather a complicated way of writing $\sec^2 x$

Harish Chandra Rajpoot
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    OP says "Does this have any nice interpretation or meaning"? The answer primarily should be yes or no, and if yes, then what it is? From the question, it is clear that he did not asked about this approach, he already knows this. –  Dec 24 '15 at 19:35