Question

Assertion :Statement 1 : ${\sin }^{-1}\left(\frac{1}{\sqrt{e}}\right)>{\tan }^{-1}\left(\frac{1}{\sqrt{\pi }}\right)$ Reason: Statement 2 : ${\sin }^{-1}x>{\tan }^{-1}y$ for $x>y,\forall x,yϵ(0,1)$

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1

Let $f\left(x\right)=si{n}^{-1}x-ta{n}^{-1}x$ , $x\ge 0$
$f\left(0\right)=0$
$f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}$
$f^{\prime }\left(x\right)>0$, for $x>0$
Hence, $si{n}^{-1}x>ta{n}^{-1}x$ for $x>0$
Now,
$e<\pi$
$\Rightarrow \frac{1}{e}>\frac{1}{\pi }$
$\Rightarrow si{n}^{-1}\left(\frac{1}{\sqrt{e}}\right)>ta{n}^{-1}\left(\frac{1}{\sqrt{\pi }}\right)$
Hence, STATEMENT 1 is TRUE.
STATEMENT 2 is also TRUE and it explains STATEMENT 1.
Hence, option A is correct.