Question

$sin^{-1}\left(\frac{1}{\sqrt{e}}\right)> tan^{-1}\left(\frac{1}{\sqrt{\pi}}\right)$ $sin^{-1}\,x>tan^{-1}\,y$ for $x>y, \forall \,x, y \,\in\left(0, 1\right)$

• A. Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
• B. Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
• C. Statement -1 is false, Statement-2 is true
• D. Statement -1 is true, Statement-2 is false

Answer 1

Answer: (A)

Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1

Answer 2

$sin^{-1}\,x = tan^{-1} \frac{x}{\sqrt{1-x^{2}}}> tan ^{-1}\,x > tan^{-1}\,y$ $\therefore$ statement-2 is true $e \frac{1}{\sqrt{\pi}}$ by statement-2. $sin^{-1}\left(\frac{1}{\sqrt{e}}\right)> tan^{-1}\left(\frac{1}{\sqrt{e}}\right) >tan^{-1} \left(\frac{1}{\sqrt{\pi}}\right)$ statement-1 is true