If cosec (A + B) = 1 and cosec(A – B) = 2, 0°

As we know that,

$\operatorname{cosec} 90^{\circ}=1$

Thus,

if $\operatorname{cosec}(A+B)=1$

$\Rightarrow A+B=90^{\circ} \quad \ldots(1)$

and $\operatorname{cosec} 30^{\circ}=2$

Thus,

if $\operatorname{cosec}(A-B)=2$

$\Rightarrow A-B=30^{\circ} \quad \ldots(2)$

Solving $(1)$ and $(2)$, we get

$A=60^{\circ}$ and $B=30^{\circ} \quad \ldots(3)$

$(\mathrm{i}) \sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$

$\cos 60^{\circ}=\frac{1}{2}$

$\sin 30^{\circ}=\frac{1}{2}$

On substituting these values, we get

$\sin A \cos B+\cos A \sin B=\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$

$=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}$

$=\frac{3}{4}+\frac{1}{4}$

$=\frac{4}{4}$

$=1$

Hence, $\sin A \cos B+\cos A \sin B=1$.

$(\mathrm{ii}) \tan 60^{\circ}=\sqrt{3}$

$\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

On substituting these values, we get

$\frac{\tan A-\tan B}{1+\tan A \tan B}=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$

$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)}$

$=\frac{\frac{3-1}{\sqrt{3}}}{1+1}$

$=\frac{2}{2 \sqrt{3}}$

$=\frac{1}{\sqrt{3}}$

Hence, $\frac{\tan A-\tan B}{1+\tan A \tan B}=\frac{1}{\sqrt{3}} .$