# Proving \frac{1}{(\tan(\frac x2)+1)^2 \cos^2(\frac x2)}=\frac{1}{1+\sin x}

Proving $\frac{1}{{\left(\mathrm{tan}\left(\frac{x}{2}\right)+1\right)}^{2}{\mathrm{cos}}^{2}\left(\frac{x}{2}\right)}=\frac{1}{1+\mathrm{sin}x}$
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Barbara Meeker
$\frac{1}{{\left(\mathrm{tan}\frac{x}{2}+1\right)}^{2}{\mathrm{cos}}^{2}\frac{x}{2}}=\frac{1}{\left({\mathrm{tan}}^{2}\frac{x}{2}+2\mathrm{tan}\frac{x}{2}+1\right){\mathrm{cos}}^{2}\frac{x}{2}}$
$=\frac{1}{{\mathrm{sin}}^{2}\frac{x}{2}+2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}+{\mathrm{cos}}^{2}\frac{x}{2}}=\frac{1}{1+2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}}=\frac{1}{1+\mathrm{sin}x}$
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Navreaiw
Let $\mathrm{tan}\frac{x}{2}=t$
So we have
$\frac{1+{t}^{2}}{{\left(t+1\right)}^{2}}=\dots =\frac{1}{1+\frac{2t}{1+{t}^{2}}}$
Now use Weierstrass substitution