If A = 45° and B = 30°, verify that $\dfrac{\text{sin A}}{\text{cos A + sin A sin B}} = \dfrac{2}{3}$.

To verify,

$\dfrac{\text{sin A}}{\text{cos A + sin A sin B}} = \dfrac{2}{3}$.

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

$\Rightarrow \dfrac{\text{sin 45°}}{\text{cos 45° + sin 45° sin 30°}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}\Big(1 + \dfrac{1}{2}\Big)} \\[1em] \Rightarrow \dfrac{\bcancel{\dfrac{1}{\sqrt{2}}}}{\bcancel{\dfrac{1}{\sqrt{2}}}\Big(\dfrac{2 + 1}{2}\Big)} \\[1em] \Rightarrow \dfrac{1}{\dfrac{3}{2}} \\[1em] \Rightarrow \dfrac{2}{3}.$

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

Hence, proved that $\dfrac{\text{sin A}}{\text{cos A + sin A sin B}} = \dfrac{2}{3}$.