# If a,b,c are positives such that a+b+c=\frac{\pi}{2} and \cot(a),\cot(b),\cot(c) is

If a,b,c are positives such that $a+b+c=\frac{\pi }{2}$ and $\mathrm{cot}\left(a\right),\mathrm{cot}\left(b\right),\mathrm{cot}\left(c\right)$ is in arithmetic progression, find $\mathrm{cot}\left(a\right)\mathrm{cot}\left(c\right)$
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Dakota Cunningham
As
$\mathrm{cot}\left(x+y\right)=\frac{\mathrm{cot}x\mathrm{cot}y-1}{\mathrm{cot}x+\mathrm{cot}y}$
$\mathrm{cot}\left(a+c\right)=\mathrm{cot}\left(\frac{\pi }{2}-b\right)=\mathrm{tan}b=\frac{1}{\mathrm{cot}b}=\frac{\mathrm{cot}a\mathrm{cot}c-1}{\mathrm{cot}a+\mathrm{cot}c}$
$\frac{1}{\mathrm{cot}b}=\frac{\mathrm{cot}a\mathrm{cot}c-1}{2\mathrm{cot}b}$
So,
$\mathrm{cot}a\mathrm{cot}c=1+2=3$
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sjkuzy5
Write A=2a etc.
$\mathrm{cot}\frac{A}{2}=\sqrt{\frac{s\left(s-a\right)}{\left(s-b\right)\left(s-c\right)}}=\frac{s\left(s-a\right)}{\mathrm{△}}$
We have $s-a+s-c=2\left(s-b\right)$
$a+c=2b⇒2s=a+b+c=3b$