Find the value of $\sin 15^{\circ}$.

Suppose $\theta_1$ and $\theta_2$ are such that $(\theta_1-\theta_2)$ lies in the $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant. If $\sin \theta_1+\sin \theta_2=-\frac{21}{65}$ and $\cos \theta_1+\cos \theta_2=-\frac{27}{65}$,then $\cos \left(\frac{\theta_1-\theta_2}{2}\right)=$

The value of $\sin \left(\frac{5 \pi}{24}\right) \cdot \cos \left(\frac{\pi}{24}\right)$ is

Let $\cos(\alpha+\beta)=-\frac{1}{10}$ and $\sin(\alpha-\beta)=\frac{3}{8}$ where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha=\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$,where $r, s \in N$,then $r+s$ is equal to . . . . . .

For $\theta \in \left(0, \frac{\pi}{2}\right)$,if $\tan 3\theta \cdot \tan 2\theta \cdot \tan \theta + \tan 2\theta + \tan \theta = 1$,then $\theta =$

The value of $\cos(18^{\circ}-A) \cos(18^{\circ}+A) - \cos(72^{\circ}-A) \cos(72^{\circ}+A)$ is equal to