If $\tan A = -\frac{1}{2}$ and $\tan B = -\frac{1}{3},$ then $A + B = $

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 . . . . . .

If $\frac{\pi}{2} < \alpha < \pi$ and $\pi < \beta < \frac{3\pi}{2}$,with $\sin \alpha = \frac{15}{17}$ and $\tan \beta = \frac{12}{5}$,then the value of $\sin(\beta - \alpha)$ is (in $/221$)

$2 \cosh (x+y) \sinh (x-y) + \sinh 2y =$

If two angles $\alpha, \beta$ are such that $0 < \alpha, \beta < \frac{\pi}{4}$,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\frac{\sqrt{1-\cos 2 \beta}}{\sqrt{1+\cos 2 \beta}}=\frac{1}{7}$,then $(2 \alpha+\beta)=$

$\tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ = $