Question

${\tan }^{-1}\left(\frac{1}{1+1+1^2}\right)+{\tan }^{-1}\left(\frac{1}{1+2+2^2}\right)+\dots +{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)=$

• A. ${\tan }^{-1}(n+1)+\pi$
• B. ${\tan }^{-1}(n+1)+\frac{7\pi }{4}$
• C. $\tan \left(n\right)(n+1)$
• D. $\tan (n+1)-\frac{\pi }{4}$

Answer 1

The correct option is D $\tan (n+1)-\frac{\pi }{4}$

Using $ta{n}^{-1}\left(\frac{A-B}{1+AB}\right)$ $=$$ta{n}^{-1}\left(A\right)$ - $ta{n}^{-1}\left(B\right)$

Given sum $=$ $ta{n}^{-1}\left(\frac{1}{1+1+1^2}\right)$ + $ta{n}^{-1}\left(\frac{1}{1+2+2^2}\right)$ +.........+ $ta{n}^{-1}\left(\frac{1}{1+n+n^2}\right)$

$=$ $ta{n}^{-1}\left(\frac{1+1-1}{1+1(1+1)}\right)$ + $ta{n}^{-1}\left(\frac{2+1-2}{1+2(1+2)}\right)$ +.........+ $ta{n}^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)$

$=$ $ta{n}^{-1}\left(2\right)$ - $ta{n}^{-1}\left(1\right)$ + $ta{n}^{-1}\left(3\right)$ - $ta{n}^{-1}\left(2\right)$ +.......+ $ta{n}^{-1}(n+1)$ - $ta{n}^{-1}\left(n\right)$

$=$ $ta{n}^{-1}(n+1)$ - $ta{n}^{-1}\left(1\right)$

$=$ $ta{n}^{-1}(n+1)$ $-$ $\frac{\pi }{4}$

Hence, option 'D' is correct.