Statement-1: If $\alpha , \beta $ are the roots of the equation 

$18(tan^{-1}x)^2 - 9 \pi tan^{-1} + \pi^2 = 0 , $ then $ \alpha + \beta = \frac{4}{\sqrt{3}}$

Statement-2: $ sec^2(cos^{-1}\frac{1}{4})+cosec^2 (sin^{-1}\frac{1}{5}) = 41 $

Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1.

We have, $(tan^{-1}x)^2 -\frac{\pi}{2}tan^{-1}x +\frac{\pi^2}{18}= 0 $

$⇒ (tan^{-1}x)^2- (\frac{\pi}{6}+\frac{\pi}{3}) tan^{-1}x +\frac{\pi^2}{18}= 0 $

$⇒ \left(tan^{-1}x - \frac{\pi}{6}\right)\left(tan^{-1}x - \frac{\pi}{3}\right)=0$

$⇒ tan^{-1} x =  \frac{\pi}{6},  \frac{\pi}{3}⇒ tan^{-1} \alpha = \frac{\pi}{6} $ and $ tan^{-1} \beta = \frac{\pi}{3}$

$⇒ \alpha = tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$ and $ \beta = tan = \frac{\pi}{3} = \sqrt{3}⇒ \alpha + \beta = \frac{4}{\sqrt{3}}$

So, statement-1 is true.

$sec^2 \left(cos^{-1}\frac{1}{4}\right) + cosec^2 \left(sin^{-1}\frac{1}{5}\right)$

$= \begin{Bmatrix} sec(sec^{-1}4 ) \end{Bmatrix}^2 + \begin{Bmatrix}cosec (cosec^{-1} 5) \end{Bmatrix}^2= 16 + 25 = 41.$

So, statement-2 is true.