For all natural numbers, find the polynomials such that cos(n theta)=p_n(tan(theta))cos^n(theta)

For all natural numbers, find the polynomials such that
$\mathrm{cos}\left(n\theta \right)={p}_{n}\left(\mathrm{tan}\left(\theta \right)\right){\mathrm{cos}}^{n}\left(\theta \right)$
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Let ${q}_{n}\left(x\right)=\left(1+ix{\right)}^{n}$
Then
${q}_{n}\left(\mathrm{tan}\left(\theta \right)\right){\mathrm{cos}}^{n}\left(\theta \right)=\left(1+i\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }{\right)}^{n}{\mathrm{cos}}^{n}\theta =\left(\mathrm{cos}\theta +i\mathrm{sin}\theta {\right)}^{n}=\mathrm{cos}n\theta +i\mathrm{sin}n\theta$
So ${p}_{n}\left(x\right)=\frac{1}{2}\left({q}_{n}\left(x\right)+{q}_{n}\left(-x\right)\right)$
Use that $\mathrm{cos}\left(-x\right)=\mathrm{cos}x$ to show that
${p}_{n}\left(\mathrm{tan}\theta \right){\mathrm{cos}}^{n}\theta =\mathrm{cos}n\theta$