Question

If $A=2ta{n}^{-1}(2\sqrt{2}-1)andB=3si{n}^{-1}\left(\frac{1}{3}\right)+si{n}^{-1}\left(\frac{3}{5}\right)$, then

• A. A = B
• B. A < B
• C. A > B
• D. None of these

Answer 1

The correct option is C

A > B

We have $A=2ta{n}^{-1}(2\sqrt{2}-1)=2ta{n}^{-1}\left(1.828\right)\Rightarrow A>2ta{n}^{-1}\sqrt{3}\Rightarrow A>\frac{2\pi }{3}$

Next $si{n}^{-1}\left(\frac{1}{3}\right)<si{n}^{-1}\left(\frac{1}{2}\right)\Rightarrow si{n}^{-1}\left(\frac{1}{3}\right)<\frac{\pi }{6}$

$\Rightarrow si{n}^{-1}\frac{1}{3}<\frac{\pi }{2}$

Also $3si{n}^{-1}\left(\frac{1}{3}\right)=si{n}^{-1}[3.\frac{1}{3}-4{\left(\frac{1}{3}\right)}^3]$

$=si{n}^{-1}\left(\frac{23}{27}\right)=si{n}^{-1}\left(0.852\right)\Rightarrow 3si{n}^{-1}\left(\frac{1}{3}\right)<si{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)\Rightarrow 3si{n}^{-1}\left(\frac{1}{3}\right)<\frac{\pi }{3}$

Further $si{n}^{-1}\left(\frac{3}{5}\right)=si{n}^{-1}\left(0.6\right)<si{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow si{n}^{-1}\left(\frac{3}{5}\right)<\frac{\pi }{3}$

Hence, $B=3si{n}^{-1}\left(\frac{1}{3}\right)+si{n}^{-1}\left(\frac{3}{5}\right)<\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}$. Hence A > B