Question

$ta{n}^{-1}n+co{t}^{-1}(n+1)$ is equal to

• A. cot ^-1 (n² - n + 1)
• B. cot ^-1 (n² + n + 1)
• C. tan ^-1 (n² + n + 1)
• D. None of these

Answer 1

The correct option is C tan ^-1 (n² + n + 1)

$ta{n}^{-1}n+co{t}^{-1}(n+1)$

$=ta{n}^{-1}+ta{n}^{-1}\frac{1}{n+1}$

$=ta{n}^{-1}\left[\frac{n+\frac{1}{n+1}}{1-n\frac{1}{n+1}}\right]$

$=ta{n}^{-1}\left[\frac{n^2+n+1}{n+1-n}\right]$

$=ta{n}^{-1}(n^2+n+1)$