Statement-1: $tan\begin{Bmatrix}cos^{-1}(\frac{1}{\sqrt{82}})-sin^{-1}(\frac{5}{\sqrt{26}})\end{Bmatrix}=\frac{29}{3}$
Statement-2: $ \begin{Bmatrix}x cos (cot^{-1}x) +sin(cot^{-1}x)\end{Bmatrix}^2 =\frac{51}{50}⇒ x=\frac{1}{5\sqrt{2}}$
Answer & explanation
Correct answer: option 4
We have, $tan\begin{Bmatrix}cos^{-1}(\frac{1}{\sqrt{82}})-sin^{-1}(\frac{5}{\sqrt{26}})\end{Bmatrix}$
$= tan \begin{Bmatrix}tan^{-1}9 - tan^{-1} 5\end{Bmatrix}= tan \begin{Bmatrix}tan^{-1} \left(\frac{9-5}{1+9×5}\right) \end{Bmatrix}$
$= tan \left(tan^{-1}\frac{2}{23}\right) = \frac{2}{23}$
So, statement-1 is not true.
$ \begin{Bmatrix}x cos (cot^{-1}x) +sin(cot^{-1}x)\end{Bmatrix}^2 =\frac{51}{50}$
$⇒ \left(\frac{x^2}{\sqrt{x^2+1}}+\frac{1}{\sqrt{x^2+1}}\right)^2 = \frac{51}{50}⇒ x^2 + 1 = \frac{51}{50} ⇒ x = \frac{1}{5\sqrt{2}}$
So, statement-2 is true.