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Question Number 155183 by mathdanisur last updated on 26/Sep/21 Answered by aleks041103 last updated on 26/Sep/21 $$\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }=\frac{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }−{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}=\frac{{cos}\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}}…
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