Question

If $A=2{\tan }^{-1}(2\sqrt{2}-1)$ and $B=3{\sin }^{-1}\frac{1}{3}+{\sin }^{-1}\frac{3}{5}$, then

• A. $A=B$
• B. $A\le B$
• C. $A>B$
• D. $A\ge B$

Answer 1

Answer: (C)

$A>B$

$A=2{\tan }^{-1}(2\sqrt{2}-1)$

Now $2\sqrt{2}-1=1.828=\sqrt{3}+x$ (approx)

So it is safe to assume that $2{\tan }^{-1}(2\sqrt{2}-1)>2{\tan }^{-1}\sqrt{3}$

$A=2{\tan }^{-1}(2\sqrt{2}-1)>\frac{2\pi }{3}$

For
$B=3{\sin }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)$

${\sin }^{-1}\frac{1}{3}<{\sin }^{-1}\frac{1}{2}$

${\sin }^{-1}\frac{1}{3}<\frac{\pi }{6}$

$3{\sin }^{-1}\frac{1}{3}<\frac{\pi }{2}$

and

${\sin }^{-1}\frac{3}{5}<{\sin }^{-1}\frac{\sqrt{3}}{2}$

${\sin }^{-1}\frac{3}{5}<\frac{\pi }{3}$

$\therefore B<\frac{\pi }{3}+\frac{\pi }{2}\Rightarrow B<\frac{5\pi }{6}$

Now,
$3{\sin }^{-1}\frac{1}{3}={\sin }^{-1}\left(3\cdot \frac{1}{3}-4(\frac{1}{3}{)}^3\right)={\sin }^{-1}\left(0.52\right)$

$\therefore 3{\sin }^{-1}\frac{1}{3}<{\sin }^{-1}\frac{\sqrt{3}}{2}$

$\therefore 3{\sin }^{-1}\frac{1}{3}<\frac{\pi }{3}$
$B<\frac{\pi }{3}+\frac{\pi }{3}$

$B<\frac{2\pi }{3}$

Thus $A>B$