Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 175953 by rexford last updated on 10/Sep/22

Answered by Ar Brandon last updated on 10/Sep/22

$$\theta =\arctan \left(\frac{1}{n^2+n+1}\right)=\arctan \left(\frac{(n+1)-n}{1+n(n+1)}\right)$$$$\tan \theta =\frac{(n+1)-n}{1+n(n+1)}=\frac{\tan \left(\arctan (n+1)\right)-\tan \left(\arctan \left(n\right)\right)}{1+\tan \left(\arctan (n+1)\right)\tan \left(\arctan \left(n\right)\right)}$$$$\Rightarrow \tan \theta =\tan ((\arctan (n+1)-\arctan \left(n\right))$$$$\Rightarrow \theta =\arctan (n+1)-\arctan \left(n\right)$$$$\Rightarrow \underset{n=1}{\sum ^m}{\theta }_n=\underset{n=1}{\sum ^m}(\arctan (n+1)-\arctan \left(n\right))$$$$=\arctan (m+1)-\arctan \left(1\right)$$$$=\arctan (m+1)-\frac{\pi }{4}$$

Commented by rexford last updated on 10/Sep/22

$$thankyouverymuch$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com