Question

$co{t}^{-1}\left({2.1}^2\right)+co{t}^{-1}\left({2.2}^2\right)+co{t}^{-1}\left({2.3}^2\right)+\cdots +$ up to $\infty$ is equal to

• A. π/4
• B. π/3
• C. π/2
• D. π/5

Answer 1

The correct option is A

π/4

$co{t}^{-1}\left({2.1}^2\right)+co{t}^{-1}\left({2.2}^2\right)+co{t}^{-1}\left({2.3}^2\right)+\cdots +\infty$
$={\sum }_{r-1}^{\infty }co{t}^{-1}\left(2.r^2\right)={\sum }_{r-1}^{\infty }ta{n}^{-1}\left(\frac{1}{2r^2}\right)$
$={\sum }_{r-1}^{\infty }ta{n}^{-1}\left(\frac{(1+2r)+(1-2r)}{1-(1+2r)(1-2r)}\right)$
$={\sum }_{r-1}^{\infty }\left[ta{n}^{-1}\right(1+2r)+ta{n}^{-1}(1-2r\left)\right]$
$=ta{n}^{-1}3-ta{n}^{-1}1+ta{n}^{-1}5-ta{n}^{-1}3+ta{n}^{-1}7-ta{n}^{-1}5+\cdots +ta{n}^{-1}\infty$
$=-\frac{\pi }{4}+\frac{\pi }{2}+=\frac{\pi }{4}$