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 $
Answer & explanation
Correct answer: option 2
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.