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 $\alpha <\beta <\gamma$.