Question

If $\alpha =2ta{n}^{-1}(2\sqrt{2}-1)$ and $\beta =3si{n}^{-1}\left(1/3\right)+si{n}^{-1}\left(3/5\right)$ then

• A. $\alpha <\beta$
• B. $\alpha =\beta$
• C. $\alpha >\beta$
• D. none of these

Answer 1

The correct option is C $\alpha >\beta$

$2{\tan }^{-1}(2\sqrt{2}-1)\Rightarrow 2{\tan }^{-1}\sqrt{3}>2\frac{\pi }{3}$
As $(2\sqrt{2}-1=1.82$ and $\sqrt{3}=1.73)$
And $3{\sin }^{-1}\frac{1}{3}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}(3\times \frac{1}{3}-4{\left(\frac{1}{3}\right)}^3)+{\sin }^{-1}\frac{3}{5}$

$=\sin \frac{3}{5}(1-\frac{4}{27})+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\frac{23}{27}+{\sin }^{-1}\frac{3}{5}<{\sin }^{-1}\frac{\sqrt{3}}{2}+si{n}^{-1}\frac{\sqrt{3}}{2}<\frac{2\pi }{3}$
As $(\frac{23}{27}=0.85,\frac{3}{5}=0.6$ and $\frac{\sqrt{3}}{2}=0.86)$
Hence, $\alpha >\beta$