Question

If $\alpha=2 \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}(3)$ and $\beta=3 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$, then

• A. $\alpha>\beta$
• B. $\alpha=\beta$
• C. $\alpha \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$
• D. $\alpha \in\left(\frac{3 \pi}{4}, \pi\right)$

Answer 1

Answer: (C)

$\alpha=2 \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1} 3=\tan ^{-1}\left(\frac{2\left(\frac{1}{2}\right)}{1-\frac{1}{4}}\right)+\tan ^{-1} 3=\tan ^{-1} \frac{4}{3}+\tan ^{-1} 3=\pi-\tan ^{-1}\left(\frac{13}{9}\right)$
$\Rightarrow \alpha \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$ and $\alpha<\beta$, as $\beta=3\left(\frac{\pi}{3}\right)+\frac{\pi}{4}=\frac{5 \pi}{4}$.