Solution:

Option(B) is correct.
To Find: The value of $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$
Now $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ can be written in terms of tan inverse as
$\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}=\tan ^{-1} \frac{1}{9}+\tan ^{-1} \frac{4}{5}$
Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{array}{l}
\Rightarrow \tan ^{-1} \frac{1}{9}+\tan ^{-1} \frac{4}{5}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{4}{5}}{1-\left(\frac{1}{9} \times \frac{4}{5}\right)}\right) \\
=\tan ^{-1}\left(\frac{41}{41}\right) \\
=\tan ^{-1}(1)=\frac{\pi}{4}
\end{array}$