Taking A = 60° and B = 30°, verify that

(i) $\dfrac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$

(ii) $\dfrac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$

(i) To verify,

$\dfrac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$.

Substituting value of A and B in L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{sin(A + B)}}{\text{cos A cos B}} \\[1em] \Rightarrow \dfrac{\text{sin(60° + 30°)}}{\text{cos 60° cos 30°}} \\[1em] \Rightarrow \dfrac{\text{sin 90°}}{\dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sqrt{3}}{4}} \\[1em] \Rightarrow \dfrac{4}{\sqrt{3}}.$

Substituting value of A and B in R.H.S. of the equation, we get :

$\Rightarrow \text{tan A + tan B} \\[1em] \Rightarrow \text{tan 60° + tan 30°} \\[1em] \Rightarrow \sqrt{3} + \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{3 + 1}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{4}{\sqrt{3}}.$

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

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

(ii) To verify,

$\dfrac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$

Substituting value of A and B in L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{sin(A - B)}}{\text{sin A sin B}} \\[1em] \Rightarrow \dfrac{\text{sin(60° - 30°)}}{\text{sin 60° sin 30°}} \\[1em] \Rightarrow \dfrac{\text{sin 30°}}{\text{sin 60° sin 30°}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2} \times \dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{4}} \\[1em] \Rightarrow \dfrac{4}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2}{\sqrt{3}}.$

Substituting value of A and B in R.H.S. of the equation, we get :

⇒ cot B - cot A

⇒ cot 30° - cot 60°

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

⇒ $\dfrac{2}{\sqrt{3}}$.

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

Hence, proved that $\dfrac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$.