Question

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

• A. $\alpha >\beta$
• B. $\alpha =\beta$
• C. $\alpha <\beta$
• D. $\alpha +\beta =2\pi$

Answer 1

The correct option is C

$\alpha <\beta$

Explanation for the correct option:

Given, $\alpha ={\sin }^{-1}\left(\frac{\surd 3}{2}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)$ and $\beta ={\cos }^{-1}\left(\frac{\surd 3}{2}\right)+{\cos }^{-1}\left(\frac{1}{3}\right)$

$\begin{array}{rcl}\therefore \alpha +\beta & =& {\sin }^{-1}\left(\frac{\surd 3}{2}\right)+{\cos }^{-1}\left(\frac{\surd 3}{2}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)+{\cos }^{-1}\left(\frac{1}{3}\right)\\ & =& \frac{\pi }{2}+\frac{\pi }{2}\\ & =& \pi \end{array}$

$\alpha =\frac{\pi }{3}+{\sin }^{-1}\left(\frac{1}{3}\right)<\frac{\pi }{3}+{\sin }^{-1}\left(\frac{1}{2}\right)$

$\because$$\sin \theta$ is increasing in $\left[0,\frac{\pi }{2}\right]$,

$\therefore$$\alpha <\frac{\pi }{3}+\frac{\pi }{6}=\frac{\pi }{2}$

$\Rightarrow$$\beta >\frac{\pi }{2}>\alpha$

$\Rightarrow$$\mathit{\alpha }\mathbf{}\mathbf{<}\mathbf{}\mathit{\beta }$

Hence, Option ‘C’ is Correct.