# If tan &#x2061;<!-- ⁡ --> x + tan 2 </msup> &#x2061;<!-- ⁡ --> x

If $\mathrm{tan}x+{\mathrm{tan}}^{2}x+{\mathrm{tan}}^{3}x=1$ then we have to find the value of $2{\mathrm{cos}}^{6}x-2{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{2}x$
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Alexis Fields
The condition gives
$\left(\mathrm{tan}x+1\right)\left({\mathrm{tan}}^{2}x+1\right)=2$
or
$2{\mathrm{cos}}^{2}x=\mathrm{tan}x+1.$
Thus,
$2{\mathrm{cos}}^{6}x-2{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{2}x=\frac{\left(\mathrm{tan}x+1{\right)}^{3}}{4}-\frac{\left(\mathrm{tan}x+1{\right)}^{2}}{2}+\frac{\mathrm{tan}x+1}{2}=$
$=\frac{{\mathrm{tan}}^{3}x+{\mathrm{tan}}^{2}x+\mathrm{tan}x+1}{4}=\frac{1}{2}.$

rmd1228887e
From the given condition:
$\mathrm{tan}x=\frac{1-{\mathrm{tan}}^{2}x}{1+{\mathrm{tan}}^{2}x}=\mathrm{cos}2x.$
Thus:
$2{\mathrm{cos}}^{6}x-2{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{2}x={\mathrm{cos}}^{2}x\left(2{\mathrm{cos}}^{2}x\left({\mathrm{cos}}^{2}x-1\right)+1\right)=$
${\mathrm{cos}}^{2}x\left(-\frac{1}{2}{\mathrm{sin}}^{2}2x+1\right)=\frac{1}{2}{\mathrm{cos}}^{2}x\left({\mathrm{cos}}^{2}2x+1\right)\stackrel{substitute}{=}$
$\frac{1}{2}{\mathrm{cos}}^{2}x\cdot \left({\mathrm{tan}}^{2}x+1\right)=\frac{1}{2}.$