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Let $e_k$ be the $k$th-degree elementary symmetric polynomial in $\tan\theta_1,\tan\theta_2,\tan\theta_3,\ldots$ (and if the sequence of $\theta$s is finite remember that the $k$th-degree elementary symmetric polynomial in $n<k$ variables is $0$). Then $$ (e_0-e_2+e_4-\cdots)^2 + (e_1-e_3+e_5-\cdots)^2 = \sec^2\theta_1\sec^2\theta_2\sec^2\theta_3\cdots $$ I haven't seen this listed among Pythagorean trigonometric identities anywhere, but maybe that means I haven't seen the sources where they would be.

So my question is: Is this "known" in the sense of being found in any authoritative or other published source?

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I don't have enough reputation to comment, but for what it's worth, if we replace $\tan(\theta_j)$ with $\lambda_j$ and set $P(x)=\prod (x-\lambda_j)$, then the left hand side will be $P(i)P(-i)$ by difference of squares and Vieta's formula, and the right hand side $\prod (1+\lambda_j^2)$. The equality then follows from $(i-\lambda_j)(-i-\lambda_j)=1+\lambda_j^2$. In this case, all that was used was $\tan^2(\theta)+1=\sec^2(\theta)$.

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    $\begingroup$ You seem to be using "$i$" both as the imaginary unit and as the index in $\theta_i$ and $\lambda_i$. $\endgroup$
    – Michael Hardy
    Jan 7, 2013 at 16:50
  • $\begingroup$ I had concluded that, like the special case $1+\tan^2\theta=\sec^2\theta$ and unlike the formula for the tangent of a sum, this identity should not depend on the fact that the circle is parametrized by arc length. But then I proved it by relying on the identity $\sec(\alpha+\beta)=\sec\alpha\sec\beta/(1-\tan\alpha\tan\beta)$. So this posting confirms that I didn't need to do it that way. $\endgroup$
    – Michael Hardy
    Jan 10, 2013 at 15:53

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