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Q.
If $\alpha=sin^{-1} \frac{\sqrt{3}}{2}+sin^{-1} \frac{1}{3}$ and $\beta=cos^{-1} \frac{\sqrt{3}}{2}+cos^{-1} \frac{1}{3},$ then :
Inverse Trigonometric Functions
Solution:
We have
$\alpha+\beta=\left(sin^{-1} \frac{\sqrt{3}}{2}+cos^{-1} \frac{\sqrt{3}}{2}\right)+\left(sin^{-1} \frac{1}{3}+cos^{-1} \frac{1}{3}\right)$
$=\frac{\pi}{2}+\frac{\pi}{2}=\pi$
Since $sin^{-1}x+cos^{-1}x=\frac{\pi}{2}$ for all $x$
Also, $\alpha=\frac{\pi}{3}+sin^{-1} \frac{1}{3} < \frac{\pi}{3}+sin^{-1} \frac{1}{3}$
as $sin \,\theta$ is increasing in $\left[0, \frac{\pi}{2}\right]$
$\therefore \alpha< \frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2} \Rightarrow \beta> \frac{\pi}{2 }> \alpha \Rightarrow \alpha < \beta$