Question

The decreasing order of $A={\sin }^{-1}\left(\frac{1-\sqrt{3}}{2\sqrt{2}}\right)$ , $B={\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ , $C={\tan }^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$ is

• A. B,A,C
• B. B,C,A
• C. C,A,B
• D. C,B,A

Answer 1

Answer: (B)

B,C,A

$A=si{n}^{-1}\left(\frac{1-\sqrt{3}}{2\sqrt{2}}\right)$
We have,$\frac{1-\sqrt{3}}{2\sqrt{2}}<0$
$\Rightarrow si{n}^{-1}\left(\frac{1-\sqrt{3}}{2\sqrt{2}}\right)<0$
Hence, $A<0$
$B=co{s}^{-1}(-\frac{1}{\sqrt{2}})=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$
$C=ta{n}^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$
We have, $\frac{\sqrt{3}+1}{2\sqrt{2}}<1$
$ta{n}^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)<\frac{\pi }{4}$
$C<\frac{\pi }{4}$
Hence, $B>C>A$
Hence, option B is correct.