Inverse Trigonometric Functions Question 84
Question: The sum of the infinite series $ {{\cot }^{-1}}2+{{\cot }^{-1}}8+{{\cot }^{-1}}18+{{\cot }^{-1}}32+… $ is,
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Let $ {S_{\infty }}={{\cot }^{-1}}2+{{\cot }^{-1}}8+{{\cot }^{-1}}18+{{\cot }^{-1}}32+… $
$ \therefore T_{n}={{\cot }^{-1}}2n^{2}={{\tan }^{-1}}\frac{1}{2n^{2}} $
$ ={{\tan }^{-1}}( \frac{2}{4n^{2}} )={{\tan }^{-1}}( \frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)} ) $
$ ={{\tan }^{-1}}(2n+1)-ta{n^{-1}}(2n-1) $
$ \therefore S_{n}=\sum\limits_{n=1}^{\infty }{{ta{n^{-1}}(2n+1)-{{\tan }^{-1}}(2n-1)}} $
$ ={{\tan }^{-1}}\infty -{{\tan }^{-1}}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4} $
BETA
