$\alpha, \beta$ are the roots of the equation $\sin ^2 x+b \sin x+c=0$. If $\alpha+\beta=\frac{\pi}{2}$, then $b^2-1=$

Suppose, $\theta_{1}$ and $\theta_{2}$ are such that $\left(\theta_{1}-\theta_{2}\right)$ lies in 3rd or 4th quadrant. If $\sin \theta_{1}+\sin \theta_{2}=-\frac{21}{65}$ and $\cos \theta_{1}+\cos \theta_{2}=-\frac{27}{65}$, then $\cos \left(\frac{\theta_{1}-\theta_{2}}{2}\right)=$

If $A$ is the solution set of the equation $\cos ^{2} x=\cos ^{2} \frac{\pi}{6}$ and $B$ is the solution set of the equation $\cos ^{2} x=\log _{16} P$ where, $P+\frac{16}{P}=10$, then, $B-A=$