Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

$f_n\left(x\right)=\sum _{j=1}^n{\tan }^{-1}\left(\frac{\left(x+j\right)-\left(x+j-1\right)}{1+\left(x+j\right)\left(x+j-1\right)}\right)$$f_n\left(x\right)=\sum _{j=1}^n\left[{\tan }^{-1}\left(x+j\right)-{\tan }^{-1}\left(x+j-1\right)\right]$
$f_n\left(x\right)={\tan }^{-1}\left(x+n\right)-{\tan }^{-1}x$
$\therefore$$\tan \left(f_n\left(x\right)\right)=\tan \left[{\tan }^{-1}\left(x+n\right)-{\tan }^{-1}x\right]$

$\tan \left(f_n\left(x\right)\right)=\tan \left[{\tan }^{-1}\frac{\left(x+n\right)-x}{1+x\left(x+n\right)}\right]$
$\tan \left(f_n\left(x\right)\right)=\frac{n}{1+x^2+nx}$,

so that $\underset{x\to \infty }{\lim }\tan \left(f_n\left(x\right)\right)=0$
$\therefore$ ${\sec }^2\left(f_n\left(x\right)\right)=1+{\tan }^2\left(f_n\left(x\right)\right)$
${\sec }^2\left(f_n\left(x\right)\right)=1+{\left(\frac{n}{1+x^2+nx}\right)}^2$
$\underset{x\to \infty }{\lim }{\sec }^2\left(f_n\left(x\right)\right)=\underset{x\to \infty }{\lim }1+{\left(\frac{n}{1+x^2+nx}\right)}^2=1$

Note that since $0$ is not in domain so first two options are wrong