What is x, if \(\displaystyle{{\cot}^{{-{1}}}{\left({3}{x}+{\frac{{{2}}}{{{x}}}}\right)}}+{{\cot}^{{-{1}}}{\left({6}{x}+{\frac{{{2}}}{{{x}}}}\right)}}+{{\cot}^{{-{1}}}{\left({10}{x}+{\frac{{{2}}}{{{x}}}}\right)}}+\cdots={1}\)

Kayden Chandler

Kayden Chandler

Answered question

2022-03-12

What is x, if cot1(3x+2x)+cot1(6x+2x)+cot1(10x+2x)+=1

Answer & Explanation

Quiensesoipsyc8n

Quiensesoipsyc8n

Beginner2022-03-13Added 3 answers

We have
cot1(A)cot1(B)=cot1(AB+1BA)
In your case this gives
cot1(i(i+1)x2+2x)=cot1(ix2)cot1((i+1)x2)
So the sum is telescopic and we get x=cot(1)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?