If a,b,c are positives such that a+b+c=π2 and cot⁡(a),cot⁡(b),cot⁡(c) is in arithmetic progression, find cot⁡(a)cot⁡(c)

As cot⁡(x+y)=cot⁡xcot⁡y−1cot⁡x+cot⁡y cot⁡(a+c)=cot⁡(π2−b)=tan⁡b=1cot⁡b=cot⁡acot⁡c−1cot⁡a+cot⁡c 1cot⁡b=cot⁡acot⁡c−12cot⁡b So, cot⁡acot⁡c=1+2=3

Write A=2a etc. cot⁡A2=s(s−a)(s−b)(s−c)=s(s−a)△ We have s−a+s−c=2(s−b) a+c=2b⇒2s=a+b+c=3b cot⁡A2cot⁡C2=ss−b=3 as 3b=2s

As cot⁡(x+y)=cot⁡xcot⁡y−1cot⁡x+cot⁡y cot⁡(a+c)=cot⁡(π2−b)=tan⁡b=1cot⁡b=cot⁡acot⁡c−1cot⁡a+cot⁡c 1cot⁡b=cot⁡acot⁡c−12cot⁡b So, cot⁡acot⁡c=1+2=3

Write A=2a etc. cot⁡A2=s(s−a)(s−b)(s−c)=s(s−a)△ We have s−a+s−c=2(s−b) a+c=2b⇒2s=a+b+c=3b cot⁡A2cot⁡C2=ss−b=3 as 3b=2s

Key to "If a,b,c are positives such that a+b+c=π2 and..."

Miguel Davenport

2022-01-24

If a,b,c are positives such that

a + b + c = \frac{π}{2}

and

\cot (a), \cot (b), \cot (c)

is in arithmetic progression, find

\cot (a) \cot (c)

Dakota Cunningham

Beginner2022-01-25Added 9 answers

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

\cot (a + c) = \cot (\frac{π}{2} - b) = \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

sjkuzy5

Beginner2022-01-26Added 11 answers

Write A=2a etc.

\cot \frac{A}{2} = \sqrt{\frac{s (s - a)}{(s - b) (s - c)}} = \frac{s (s - a)}{△}

We have

s - a + s - c = 2 (s - b)

a + c = 2 b \Rightarrow 2 s = a + b + c = 3 b

\cot \frac{A}{2} \cot \frac{C}{2} = \frac{s}{s - b} = 3 as 3 b = 2 s

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