If A = 60° and B = 30°, prove that :

(i) sin (A + B) = sin A cos B + cos A sin B

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

(iii) cos (A - B) = cos A cos B + sin A sin B

(iv) tan (A - B) = $\dfrac{\tan A - \tan B}{1 + \tan A \tan B}$

(i) Left Hand Side

sin (A + B) = sin (60° + 30°) = sin 90°

= 1

Right Hand Side

sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

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

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

= 1

Hence, proved that sin (A + B) = sin A cos B + cos A sin B.

(ii) Left Hand Side

cos (A + B) = cos (60° + 30°) = cos 90°

= 0

Right Hand Side

cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

= $\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}$

= $\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}$

= 0.

Hence, proved that cos (A + B) = cos A cos B - sin A sin B.

(iii) Left Hand Side

cos (A - B) = cos (60° - 30°) = cos 30°

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

Right Hand Side

cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

$\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}\\[1em] = \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{4}\\[1em] = \dfrac{2\sqrt{3}}{4}\\[1em] =\dfrac{\sqrt{3}}{2}$

Hence, proved that cos (A - B) = cos A cos B + sin A sin B.

(iv) Left Hand Side :

tan (A - B) = tan (60° - 30°) = tan 30°

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

Right Hand Side

$\Rightarrow \Big(\dfrac{\tan A - \tan B}{1 + \tan A \tan B}\Big)\\[1em] = \Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)\\[1em] = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\dfrac{1}{\sqrt{3}}}\\[1em] =\dfrac{\dfrac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\dfrac{\dfrac{2}{\sqrt{3}}}{2}\\[1em] = \dfrac{1}{\sqrt{3}}.$

Hence, proved that tan (A - B) = $\dfrac{\tan A - \tan B}{1 + \tan A \tan B}$.