Question

Let $\alpha=\cot ^{-1}\left(\frac{\pi}{3}\right), \beta=\sin ^{-1}\left(\frac{\pi}{4}\right)$ and $\gamma=\sec ^{-1}\left(\frac{2 \pi}{3}\right)$, then the correct order sequence is

• A. $\alpha<\gamma<\beta$
• B. $\beta<\alpha<\gamma$
• C. $\gamma<\beta<\alpha$
• D. $\alpha<\beta<\gamma$

Answer 1

Answer: (D)

As $\frac{\pi}{3}>1 \Rightarrow \cot ^{-1} \frac{\pi}{3}<\cot ^{-1} 1=\frac{\pi}{4} \Rightarrow \cot ^{-1} \frac{\pi}{3}<\frac{\pi}{4}=0.7856$
As $ \frac{\pi}{4}>\frac{1}{\sqrt{2}} \Rightarrow \sin ^{-1} \frac{\pi}{4}>\sin ^{-1} \frac{1}{\sqrt{2}}=\frac{\pi}{4} \Rightarrow \sin ^{-1} \frac{\pi}{4}>\frac{\pi}{4}=0.7856$
As $ \frac{2 \pi}{3}>2 \Rightarrow \sec ^{-1} \frac{2 \pi}{3}>\sec ^{-1} 2=\frac{\pi}{3} \Rightarrow \sec ^{-1} \frac{2 \pi}{3}>\frac{\pi}{3} \simeq 1.0476$
Clearly $\alpha<\beta<\gamma$.