# Find the exact value of cot &#x2061;<!--

Find the exact value of $\mathrm{cot}\left({202.5}^{\circ }\right)$
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haingear8v
$\mathrm{cot}\left({202.5}^{0}\right)=\mathrm{cot}\left({180}^{0}+{22.5}^{0}\right)=\mathrm{cot}\left({22.5}^{0}\right)$
Now Calculate value of $\mathrm{tan}\left({22.5}^{0}\right)$
Using
$\mathrm{tan}\left(2A\right)=\frac{2\mathrm{tan}A}{1-{\mathrm{tan}}^{2}A}\phantom{\rule{thickmathspace}{0ex}},$
Put $A={22.5}^{0}$
We get
$\mathrm{tan}\left({45}^{0}\right)=\frac{2\mathrm{tan}\left({22.5}^{0}\right)}{1-{\mathrm{tan}}^{2}\left({22.5}^{0}\right)}⇒2\mathrm{tan}\left({22.5}^{0}\right)=1-{\mathrm{tan}}^{2}\left({22.5}^{0}\right)$
So we get
${\mathrm{tan}}^{2}\left({22.5}^{0}\right)+2\mathrm{tan}\left({22.5}^{0}\right)+1=2⇒{\left[1+\mathrm{tan}\left({22.5}^{0}\right)\right]}^{2}=\left(\sqrt{2}{\right)}^{2}$
So we get
$1+\mathrm{tan}\left({22.5}^{0}\right)=±\sqrt{2}$
So we get
$\mathrm{tan}\left({22.5}^{0}\right)=\sqrt{2}-1$
bcz $\mathrm{tan}\left({22.5}^{0}\right)>0$
So we get
$\mathrm{cot}\left({22.5}^{0}\right)=\frac{1}{\mathrm{tan}\left({22.5}^{0}\right)}=\frac{1}{\sqrt{2}-1}×\frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$
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Cristopher Knox
$\mathrm{cot}{202.5}^{0}=\mathrm{cot}22.5$ , in third quadrant. Recognizing half of ${45}^{0}$ you are going to need tan of half angle formula.
Denoting tan full angle by caps, $T=\frac{2t}{1-{t}^{2}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}t=\left(\sqrt{1+T}-1\right),\mathrm{tan}{22.5}^{0}=\sqrt{2}-1$ where we chose positive value from $±$ in third quadrant...and the required cot is its reciprocal:
$\mathrm{cot}{22.5}^{0}=\sqrt{2}+1.$