<s>[INST] 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 $\cot(a)\cot(c)$ 
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.
 [/INST] 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,

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

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