If $a,b,c$ are positives such that $a+b+c=\pi/2$ and $\cot(a),\cot(b),\cot(c)$ is in arithmetic progression, find $\cot(a)\cot(c)$

Question

Suppose $a,b,c$ are positive reals such that $a+b+c=\pi/2$ and $2\cot(b)=\cot(a)+\cot(c)$.

Find $\cot(a) \cdot \cot(c)$.

I have tried writing $b=\pi/2-a-c$ into the given equation and play with trigonometric identities, but it just got too messy and I don't know what else to try.

Answer 1

As

$$\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$

$$\cot(a+c) = \cot\left(\frac{\pi}{2}-b\right) = \tan(b) = \frac{1}{\cot b}= \frac{\cot a\cot c -1}{\cot a+ \cot c}$$

$$\frac{1}{\cot b} = \frac{\cot a\cot c - 1}{2\cot b}$$

So,

$$\cot a \cot c = 1+2 = 3$$

Answer 2

Using https://www.askiitians.com/iit-jee-properties-and-solutions-of-triangles/solution-of-triangles/

Write $A=2a$ etc.

$\cot\dfrac A2=\sqrt{\dfrac{s(s-a)}{(s-b)(s-c)}}=\dfrac{s(s-a)}{\triangle}$

We have $s-a+s-c=2(s-b)$

$a+c=2b\implies2s=a+b+c=3b$

$\cot\dfrac A2\cot\dfrac C2=\dfrac s{s-b}=3$ as $3b=2s$