July 2023 1 43 Report
[Trigonometria: Identidades trigonométricas]

Mostre que para todo n natural ≥ 1, vale

     [tex]\dfrac{\mathrm{tg}(2^n x)}{\prod\limits_{k=1}^n[\sec(2^k x)+1]}=\mathrm{tg}(x)[/tex]

sob a condição de existência

     [tex]x\in\{t\in\mathbb{R}:~\cos(2^k t)\ne 0,\,\ne -1,~k\in\{0,\,1,\ldots,\, n\}\}.[/tex]​
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