Prove $\mathrm{sec}2x+\mathrm{tan}2x\equiv \frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\mathrm{sin}x}$

Adriana Ayers
2022-06-29
Answered

Prove $\mathrm{sec}2x+\mathrm{tan}2x\equiv \frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\mathrm{sin}x}$

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mallol3i

Answered 2022-06-30
Author has **20** answers

$\mathrm{sec}2x+\mathrm{tan}2x={\displaystyle \frac{1+\mathrm{sin}2x}{\mathrm{cos}2x}}={\displaystyle \frac{(\mathrm{cos}x+\mathrm{sin}x{)}^{2}}{(\mathrm{cos}x+\mathrm{sin}x)(\mathrm{cos}x-\mathrm{sin}x)}}=?$

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However, I am unable to see why this would be? It is probably just a simple oversight but I am unsure.

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Does the identity ${\mathrm{cos}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(x\right)=1$ hold in a unital Banach algebra where 1 is the unit?

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Let's assume that we have an unital Banach algebra T and we define sine and cosine using the normal power series definition as for

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$\frac{{\mathrm{cos}}^{2}(\theta +\alpha )}{1-{\mathrm{cos}}^{2}(\theta -\alpha )}=\text{const.}$

where$\theta$ is a variable and $\alpha$ is the number satisfying

$\alpha ={\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)$

where

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