If tan A = 1 and tan B = $\sqrt3$ ; evaluate:

(i) cos A cos B - sin A sin B.

(ii) sin A cos B + cos A sin B.

Given: tan A = 1 and tan B = $\sqrt3$

⇒ tan A = tan 45° and tan B = tan 60°

⇒ A = 45° and B = 60°

(i) cos A cos B - sin A sin B

= cos 45° cos 60° - sin 45° sin 60°

$= \dfrac{1}{\sqrt{2}} . \dfrac{1}{2} - \dfrac{1}{\sqrt{2}} . \dfrac{\sqrt{3}}{2}\\[1em] = \dfrac{1 - \sqrt{3}}{2\sqrt{2}}\\[1em] = \dfrac{\sqrt{2}(1 - \sqrt{3})}{2\sqrt{2}.\sqrt{2}}\\[1em] = \dfrac{\sqrt{2} - \sqrt{6}}{4}$

Hence, the value of cos A cos B - sin A sin B = $\dfrac{\sqrt{2} - \sqrt{6}}{4}$.

(ii) sin A cos B + cos A sin B

= sin 45° cos 60° + cos 45° sin 60°

$= \dfrac{1}{\sqrt{2}} . \dfrac{1}{2} + \dfrac{1}{\sqrt{2}} . \dfrac{\sqrt{3}}{2}\\[1em] = \dfrac{1 + \sqrt{3}}{2\sqrt{2}}\\[1em] = \dfrac{\sqrt{2}(1 + \sqrt{3})}{2\sqrt{2}.\sqrt{2}}\\[1em] = \dfrac{\sqrt{2} + \sqrt{6}}{4}$

Hence, the value of sin A cos B - cos A sin B = $\dfrac{\sqrt{2} + \sqrt{6}}{4}$.