If A = 60° and B = 30°, verify 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) sin(A - B) = sin A cos B - cos A sin B

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

(i) To verify,

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

Substituting values in L.H.S. of the above equation :

sin(A + B) = sin(60° + 30°) = sin 90° = 1.

Substituting values in R.H.S. of the equation :

sin A cos B + cos A sin B = sin 60° cos 30° + cos 60° sin 30°

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

Since, L.H.S. = R.H.S.

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

(ii) To verify,

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

Substituting values in L.H.S. of equation :

cos(A + B) = cos(60° + 30°) = cos 90° = 0.

Substituting values in R.H.S. of equation :

cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} \\[1em] = 0.$

Since, L.H.S. = R.H.S.

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

(iii) To verify,

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

Substituting values in L.H.S. of equation :

sin(A - B) = sin(60° - 30°) = sin 30° = $\dfrac{1}{2}$.

Substituting values in R.H.S. of equation :

sin A cos B - cos A sin B = sin 60° cos 30° - cos 60° sin 30°

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

Since, L.H.S. = R.H.S.

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

(iv) To verify,

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

Substituting values in L.H.S. of equation :

tan (A - B) = tan (60° - 30°) = tan 30° = $\dfrac{1}{\sqrt{3}}$.

Substituting values in R.H.S. of equation :

$\dfrac{\text{tan A - tan B}}{1 + \text{ tan A tan B}} = \dfrac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}} \\[1em] = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}} \\[1em] = \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{2} \\[1em] = \dfrac{1}{\sqrt{3}}.$

Since, L.H.S. = R.H.S.

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