Question

If $\alpha =si{n}^{-1}\frac{\sqrt{3}}{2}+si{n}^{-1}\frac{1}{3}$ and $\beta =co{s}^{-1}\frac{\sqrt{3}}{2}+co{s}^{-1}\frac{1}{3},$ then :

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

Answer 1

Answer: (A)

We have
$\alpha +\beta =\left(si{n}^{-1}\frac{\sqrt{3}}{2}+co{s}^{-1}\frac{\sqrt{3}}{2}\right)+\left(si{n}^{-1}\frac{1}{3}+co{s}^{-1}\frac{1}{3}\right)$
$=\frac{\pi }{2}+\frac{\pi }{2}=\pi$
Since $si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}$ for all $x$
Also, $\alpha =\frac{\pi }{3}+si{n}^{-1}\frac{1}{3}<\frac{\pi }{3}+si{n}^{-1}\frac{1}{3}$
as $sin\theta$ is increasing in $\left[0,\frac{\pi }{2}\right]$
$\therefore \alpha <\frac{\pi }{3}+\frac{\pi }{6}=\frac{\pi }{2}\Rightarrow \beta >\frac{\pi }{2}>\alpha \Rightarrow \alpha <\beta$