Question

Let $\alpha ,\beta (\alpha >\beta )$ be the roots of the equation ${\sin }^{-1}x-\frac{1}{{\sin }^{-1}x}={\cos }^{-1}x-\frac{1}{{\cos }^{-1}x}.$ Then

• A. $\alpha =\frac{1}{\sqrt{2}}$
• B. $\beta =-\frac{1}{\sqrt{2}}$
• C. ${\sin }^{-1}\alpha =\frac{\pi }{3}$
• D. ${\sin }^{-1}\beta =\frac{\pi }{4}-\sqrt{1+\frac{{\pi }^2}{16}}$

Answer 1

Answer: (A), (D)

${\sin }^{-1}\beta =\frac{\pi }{4}-\sqrt{1+\frac{{\pi }^2}{16}}$

${\sin }^{-1}x-\frac{1}{{\sin }^{-1}x}={\cos }^{-1}x-\frac{1}{{\cos }^{-1}x}\Rightarrow {\sin }^{-1}x-{\cos }^{-1}x+\frac{1}{{\cos }^{-1}x}-\frac{1}{{\sin }^{-1}x}=0\Rightarrow ({\sin }^{-1}x-{\cos }^{-1}x)(1+\frac{1}{{\cos }^{-1}x{\sin }^{-1}x})=0\Rightarrow ({\sin }^{-1}x-{\cos }^{-1}x)\frac{({\sin }^{-1}x{\cos }^{-1}x+1)}{\left({\sin }^{-1}x\right)\left({\cos }^{-1}x\right)}=0\Rightarrow {\sin }^{-1}x={\cos }^{-1}x$
$\Rightarrow {\sin }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x\Rightarrow {\sin }^{-1}x=\frac{\pi }{4}\Rightarrow x=\frac{1}{\sqrt{2}}$
or
${\sin }^{-1}x{\cos }^{-1}x+1=0\Rightarrow {\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x)+1=0\Rightarrow ({\sin }^{-1}x{)}^2-\frac{\pi }{2}({\sin }^{-1}x)-1=0\Rightarrow {\sin }^{-1}x=\frac{\frac{\pi }{2}\pm \sqrt{(\frac{{\pi }^2}{4}+4)}}{2}\Rightarrow {\sin }^{-1}x=\frac{\pi }{4}-\sqrt{(1+\frac{{\pi }^2}{16})}$