If

Given: $\sin A=\frac{1}{2}$ and $\cos B=\frac{\sqrt{3}}{2}$

Here, $\frac{\pi}{2}

That is, $A$ is in the second quadrant and $B$ is in the first quadrant.

We know that in the second quadrant, sine function is positive and cosine and tan functions are negative In the first quadrant, all $\mathrm{T}$ - functions are positive.

Therefore,

$\cos A=-\sqrt{1-\sin ^{2} A}=-\sqrt{1-\left(\frac{1}{2}\right)^{2}}=-\sqrt{1-\frac{1}{4}}=-\sqrt{\frac{3}{4}}=\frac{-\sqrt{3}}{2}$

$\tan A=\frac{\sin A}{\cos A}=\frac{1 / 2}{-\sqrt{3} / 2}=\frac{-1}{\sqrt{3}}$

$\sin B=\sqrt{1-\cos ^{2} A}=\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

$\tan B=\frac{\sin B}{\cos B}=\frac{1 / 2}{\sqrt{3} / 2}=\frac{1}{\sqrt{3}}$

Now,

(i) $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

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

$=\frac{0}{1+\frac{1}{3}}=0$

(ii) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

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

$=\frac{\frac{-2}{\sqrt{3}}}{1-\frac{1}{3}}$

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

$=-\sqrt{3}$