Inverse Trigonometric Functions Question 71
Question: $ \sum\limits_{r=1}^{\infty }{{{\tan }^{-1}}( \frac{1}{1+r+r^{2}} )=….} $
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{4} $
C) $ \frac{2\pi }{3} $
D) None
Show Answer
Answer:
Correct Answer: B
$ \frac{1}{1+r+r^{2}}=\frac{1}{1+r(r+1)}=\frac{\frac{1}{r(r+1)}}{1+\frac{1}{r(r+1)}}=\frac{\frac{1}{r}-\frac{1}{r+1}}{1+\frac{1}{r}( \frac{1}{r+1} )} $
$ \therefore {{\tan }^{-1}}( \frac{1}{1+r+r^{2}} )={{\tan }^{-1}}\frac{1}{r}-{{\tan }^{-1}}\frac{1}{r+1} $
$ \therefore \sum\limits_{r=1}^{\infty }{{{\tan }^{-1}}( \frac{1}{1+r+r^{2}} )}={{\tan }^{-1}}1=\frac{\pi }{4} $
BETA
