The equation of the sides OA, OB and  AB  of $\mathrm{\Delta OAB}$ are $y=0, x=0$and $\mathrm{x}+\mathrm{y}=\frac{\sqrt{3}}{\sqrt{2}}$ respectively. 
Point $\mathrm{P}=(\mathrm{sin\theta },\mathrm{cos\theta })$ will lie inside  the $\mathrm{\Delta OAB}$ . then which of the following is/are TRUE
 

P  lies inside of $\mathrm{\Delta OAB}$ if $\mathrm{sin\theta }>0,\mathrm{cos\theta }>0$ and $\mathrm{sin\theta }+\mathrm{cos\theta }<\frac{\sqrt{3}}{\sqrt{2}}$
Now, $\mathrm{sin\theta }>0,\mathrm{cos\theta }>0$
$\Rightarrow 0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2}[\therefore 0\le \mathrm{\theta }\le 2\mathrm{\pi }].....\left(1\right)$
$\begin{array}{l}\frac{1}{\sqrt{2}}\sin \mathrm{\theta }+\frac{1}{\sqrt{2}}\cos \mathrm{\theta }<\frac{\sqrt{3}}{2}\\ \sin \left(\mathrm{\theta }+\frac{\mathrm{\pi }}{4}\right)<\frac{\sqrt{3}}{2}\\ \sin \left(\mathrm{\theta }+\frac{\mathrm{\pi }}{4}\right)<\frac{\sqrt{3}}{2}\\ 0<\mathrm{\theta }+\frac{\mathrm{\pi }}{4}<\frac{\mathrm{\pi }}{3}\text{ and }\frac{2\mathrm{\pi }}{3}<\mathrm{\theta }+\frac{\mathrm{\pi }}{4}<\mathrm{\pi }\\ -\frac{\mathrm{\pi }}{4}<\mathrm{\theta }<\frac{\mathrm{\pi }}{12},\frac{5\mathrm{\pi }}{12}<\mathrm{\theta }<\frac{3\mathrm{\pi }}{4}\dots .\left(2\right)\end{array}$
FROM 1 AND 2
$\mathrm{\theta }\in (0,\frac{\mathrm{\pi }}{12})\cup (\frac{5\mathrm{\pi }}{12},\frac{\mathrm{\pi }}{2})$